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[LeetCode]606. Construct String from Binary Tree (Easy) Python-2023.12.08

Uriel — Fri, 08 Dec 2023 16:12:00 GMT

先序遍历二叉树�ƈ且格式化输出�Q�DFS
注意如果只有叛_��树、左子树为空的话也要输出左子树的一�Ҏ��?br />

1 #606
2 #Runtime: 44 ms (Beats 7.3%)
3 #Memory: 16.2 MB (Beats 35.94%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def tree2str(self, root):
13         """
14         :type root: TreeNode
15         :rtype: str
16         """
17         self.ans = ""
18         def DFS(node):
19             self.ans += str(node.val)
20             if node.left or node.right:
21                 self.ans += '('
22                 if node.left:
23                     DFS(node.left)
24                 self.ans += ')'
25                 if node.right:
26                     self.ans += '('
27                     DFS(node.right)
28                     self.ans += ')'
29
30         if root:
31             DFS(root)
32         return self.ans

Uriel 2023-12-09 00:12 发表评论

[LeetCode]515. Find Largest Value in Each Tree Row (Medium) Python-2023.10.24

Uriel — Tue, 24 Oct 2023 10:40:00 GMT

求二叉树每一层的最大��|��按层BFS

1 #515
2 #Runtime: 26 ms (Beats 68.6%)
3 #Memory: 17.7 MB (Beats 71.73%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def largestValues(self, root):
13         """
14         :type root: TreeNode
15         :rtype: List[int]
16         """
17         if not root:
18             return []
19         q = deque([root])
20         ans = []
21         while q:
22             sz = len(q)
23             t = float('-inf')
24             while sz:
25                 sz -= 1
26                 node = q.popleft()
27                 t = max(t, node.val)
28                 if node.left:
29                     q.append(node.left)
30                 if node.right:
31                     q.append(node.right)
32             ans.append(t)
33         return ans

Uriel 2023-10-24 18:40 发表评论

[LeetCode]1361. Validate Binary Tree Nodes (Medium) Python-2023.10.17

Uriel — Tue, 17 Oct 2023 14:00:00 GMT

�l�出一颗二叉树节点数量和每个节点的左儿子label集合以及叛_��子label集合�Q�问是否是一颗二叉树�Q�先用set选出root然后BFS�Q�如果同一个节点访问不止一�ơ或者有节点讉K��不到�Q�则不是二叉�?br />

1 #1361
2 #Runtime: 224 ms (Beats 92.11%)
3 #Memory: 15.5 MB (Beats 71.5%�Q?/span>
4
5 class Solution(object):
6     def validateBinaryTreeNodes(self, n, leftChild, rightChild):
7         """
8         :type n: int
9         :type leftChild: List[int]
10         :type rightChild: List[int]
11         :rtype: bool
12         """
13         rt = 0
14         child_nodes = set(leftChild + rightChild)
15         for i in xrange(n):
16             if i not in child_nodes:
17                 rt = i
18         vis = set()
19         q = deque([rt])
20         while q:
21             node = q.popleft()
22             if node in vis:
23                 return False
24             vis.add(node)
25             if leftChild[node] != -1:
26                 q.append(leftChild[node])
27             if rightChild[node] != -1:
28                 q.append(rightChild[node])
29         return len(vis) == n

Uriel 2023-10-17 22:00 发表评论

[LeetCode]847. Shortest Path Visiting All Nodes (Hard) Python-2023.06.17

Uriel — Sun, 17 Sep 2023 11:02:00 GMT

�l�出一个有向图每个节点的链接情况（graph[i]表示与节点i相连的节点）�Q�问最��走�q�多��蟩变可以遍历所有节点，BFS�Q�用二进制mask记录走过的节点，vis[bit_mask][node]=1记录已经走过bit_mask中存储的节点�Q��ƈ且最后刚刚经�q�node节点

1 #847
2 #Runtime: 76 ms (Beats 100%)
3 #Memory: 14.2 MB (Beats 87.50%)
4
5 class Solution(object):
6     def shortestPathLength(self, graph):
7         """
8         :type graph: List[List[int]]
9         :rtype: int
10         """
11         n = len(graph)
12         vis_mask = (1 << n) - 1
13         q = deque()
14         vis = [[0] * n for _ in range(vis_mask + 1)]
15         for x in xrange(n):
16             ini_mask = 1 << x
17             q.append((x, ini_mask, 1))
18             vis[ini_mask][x] = 1
19         while q:
20             t = q.popleft()
21             cur_node, cur_mask, cur_len = t
22             if cur_mask == vis_mask:
23                 return cur_len - 1
24             for nei in graph[cur_node]:
25                 new_mask = cur_mask | (1 << nei)
26                 if vis[new_mask][nei]:
27                     continue
28                 q.append((nei, new_mask, cur_len + 1))
29                 vis[new_mask][nei] = 1
30         return -1

Uriel 2023-09-17 19:02 发表评论

[LeetCode]332. Reconstruct Itinerary (Hard) Python-2023.09.14

Uriel — Thu, 14 Sep 2023 07:56:00 GMT

�l�出list tickets�Q?tickets[i][0], tickets[i][1])表示一张票的�v点和�l�点�Q�每个站点用三个字母的字�W�串表示�Q�从“JFK”站开始，历经所有站点，输出�l�过的节点的�ơ序�Q�若有不止一条线路，输出字母序最��的解，DFS�Q�注意string list的拼接方式（后面�?#8216;,’)。写法参�?-> https://leetcode.com/problems/reconstruct-itinerary/solutions/78772/python-dfs-backtracking/?envType=daily-question&envId=2023-09-14

1 #332
2 #Runtime: 62 ms (Beats 38.57%)
3 #Memory: 14.1 MB (Beats 53.57%)
4
5 class Solution(object):
6     def findItinerary(self, tickets):
7         """
8         :type tickets: List[List[str]]
9         :rtype: List[str]
10         """
11         node = defaultdict(list)
12         for (x, y) in tickets:
13             node[x] += y,
14         self.ans = ["JFK"]
15         def DFS(st):
16             if len(self.ans) == len(tickets) + 1:
17                 return self.ans
18             tp_dst = sorted(node[st])
19             for dst in tp_dst:
20                 node[st].remove(dst)
21                 self.ans += dst,
22                 ok = DFS(dst)
23                 if ok:
24                     return ok
25                 self.ans.pop()
26                 node[st] += dst,
27         return DFS("JFK")

Uriel 2023-09-14 15:56 发表评论

[LeetCode]77. Combinations (Medium) C++/Python-2014.01.18/2023.08.01

Uriel — Tue, 01 Aug 2023 12:22:00 GMT

生成1-n选k个数的所有组合，DFS

C++

1 //77
2 //Runtime 48 ms (Beats 24.56%)
3
4 class Solution {
5 public:
6     vectorint>> res;
7     int nt;
8
9     void DFS(int n, int k, vector<int> &tp, int pos) {
10         if(nt == k) {
11             res.push_back(tp);
12             return;
13         }
14         for(int i = pos; i <= n; ++i) {
15             tp.push_back(i);
16             nt++;
17             DFS(n, k, tp, i + 1);
18             tp.pop_back();
19             nt--;
20         }
21     }
22
23     vectorint> > combine(int n, int k) {
24         vector<int> tp;
25         //sort(S.begin(), S.end());
26         res.clear();
27         nt = 0;
28         DFS(n, k, tp, 1);
29         return res;
30     }
31 };

Python

1 #77
2 #Runtime: 340 ms (Beats 77%)
3 #Memory: 14.9 MB (Beats 55.68%)
4
5 class Solution(object):
6     def combine(self, n, k):
7         """
8         :type n: int
9         :type k: int
10         :rtype: List[List[int]]
11         """
12         self.ans = []
13         def dfs(tp, p):
14             if len(tp) == k:
15                 self.ans.append(tp[:])
16                 return
17             for i in range(p, n + 1):
18                 tp.append(i)
19                 dfs(tp, i + 1)
20                 tp.pop()
21         dfs([], 1)
22         return self.ans

Uriel 2023-08-01 20:22 发表评论

[LeetCode]688. Knight Probability in Chessboard (Medium) Python3-2023.07.22

Uriel — Sat, 22 Jul 2023 08:41:00 GMT

已知��盘��D��长和knight的位�|?row, column)�Q�问k步之后knight仍然在棋盘格上的概率�Q�DFS+

lru_cache

1 #688
2 #Runtime: 225 ms (Beats 75.30%)
3 #Memory: 26.5 MB (Beats 13.8%)
4
5 class Solution:
6     def knightProbability(self, n: int, k: int, row: int, column: int) -> float:
7         d = [[1, 2], [2, 1], [1, -2], [2, -1], [-1, 2], [-2, 1], [-1, -2], [-2, -1]]
8
9         @lru_cache(None)
10         def dfs(x, y, k):
11             if k == 0:
12                 return 1
13             ans = 0
14             for dx, dy in d:
15                 tx = x + dx
16                 ty = y + dy
17                 if 0 <= tx < n and 0 <= ty < n:
18                     ans += dfs(tx, ty, k - 1)
19             return ans
20         if not k:
21             return 1
22         return dfs(row, column, k) / 8**k

Uriel 2023-07-22 16:41 发表评论

[LeetCode]1751. Maximum Number of Events That Can Be Attended II (Hard) Python-2023.07.15

Uriel — Sat, 15 Jul 2023 10:04:00 GMT

有一堆event�Q�每个event有开始和�l�束旉��以及value�Q�最多参加k个event问最多获得多��value�Q�DFS+二分查找下一个可以参与的开始时间最早的event

1 #1751
2 #Runtime: 999 ms (Beats 60%)
3 #Memory: 133.9 MB (Beats 20%)
4
5 class Solution(object):
6     def maxValue(self, events, k):
7         """
8         :type events: List[List[int]]
9         :type k: int
10         :rtype: int
11         """
12         n = len(events)
13         events.sort(key=lambda x: x[0])
14         dp = [[-1] * n for _ in range(k + 1)]
15
16         def DFS(idx, cnt):
17             if cnt == 0 or idx == len(events):
18                 return 0
19             if dp[cnt][idx] != -1:
20                 return dp[cnt][idx]
21             dp[cnt][idx] = max(DFS(idx + 1, cnt), events[idx][2] + DFS(bsearch(events[idx][1]), cnt - 1))
22             return dp[cnt][idx]
23
24         def bsearch(x):
25             l, r = 0, len(events) - 1
26             while l <= r:
27                 mid = (l + r + 1) // 2
28                 if events[mid][0] <= x:
29                     l = mid + 1
30                 else:
31                     r = mid - 1
32             return l
33
34         return DFS(0, k)

Uriel 2023-07-15 18:04 发表评论

[LeetCode]207. Course Schedule (Medium) Python-2023.07.13

Uriel — Thu, 13 Jul 2023 08:25:00 GMT

一�?div style="display: inline-block;">numCourses门课�Q?div style="display: inline-block;">prerequisites中的每个pair [a, b]表示a的先修课是b�Q�问是否有可能修完所有课
BFS+拓扑排序思想

1 #207
2 #Runtime: 74 ms (Beats 75.97%)
3 #Memory: 14.5 MB (Beats 97.82%)
4
5 class Solution(object):
6     def canFinish(self, numCourses, prerequisites):
7         """
8         :type numCourses: int
9         :type prerequisites: List[List[int]]
10         :rtype: bool
11         """
12         ind = [0] * numCourses
13         adj = [[] for _ in range(numCourses)]
14         for i in range(len(prerequisites)):
15             a, b = prerequisites[i]
16             adj[b].append(a)
17             ind[a] += 1
18         q = deque()
19         for i in range(numCourses):
20             if not ind[i]:
21                 q.append(i)
22         ans = []
23         while q:
24             node = q.popleft()
25             ans.append(node)
26             for i in adj[node]:
27                 ind[i] -= 1
28                 if ind[i] == 0:
29                     q.append(i)
30         return len(ans) == numCourses

Uriel 2023-07-13 16:25 发表评论

[LeetCode]863. All Nodes Distance K in Binary Tree (Medium) Python-2023.07.11

Uriel — Tue, 11 Jul 2023 09:09:00 GMT

求二叉树中和某个target节点距离为k的所有节点��|��先BFS建图�Q�python的dict存图�Q�，然后在图上BFS求解

1 #863
2 #Runtime: 25 ms (Beats 49.62%)
3 #Memory: 13.6 MB (Beats 74.5%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, x):
8 #         self.val = x
9 #         self.left = None
10 #         self.right = None
11
12 class Solution(object):
13     def distanceK(self, root, target, k):
14         """
15         :type root: TreeNode
16         :type target: TreeNode
17         :type k: int
18         :rtype: List[int]
19         """
20         graph = {}
21         dep_target = 0
22         q = deque([[root, -1]])
23         while q:
24             node, f = q.popleft()
25             if node.val not in graph:
26                 if f != -1:
27                     graph[node.val] = [f]
28                     graph[f].append(node.val)
29                 else:
30                     graph[node.val] = []
31             else:
32                 graph[node.val].append(f)
33                 graph[f].append(node.val)
34             if node.left:
35                 q.append([node.left, node.val])
36             if node.right:
37                 q.append([node.right, node.val])
38         q = deque([target.val])
39         dis = 0
40         ans = []
41         vis = set([target.val])
42         while q and dis < k:
43             sz = len(q)
44             while sz:
45                 sz -= 1
46                 node = q.popleft()
47                 for i in graph[node]:
48                     if i not in vis:
49                         q.append(i)
50                         vis.add(i)
51             dis += 1
52         if dis == k:
53             ans = list(q)
54         return ans

Uriel 2023-07-11 17:09 发表评论

[LeetCode]111. Minimum Depth of Binary Tree (Easy) C++/Python-2014.01.18/2021.01.25/2023.07.10

Uriel — Mon, 10 Jul 2023 05:58:00 GMT

求二叉树的最��深度，BFS或者DFS

C++ BFS

1 //111
2 //Runtime: 32 ms (Beats 100%)
3
4 /**
5  * Definition for binary tree
6  * struct TreeNode {
7  *     int val;
8  *     TreeNode *left;
9  *     TreeNode *right;
10  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
11  * };
12  */
13 class Solution {
14 public:
15     struct Que {
16         int depth;
17         TreeNode *p;
18     }q[100010];
19
20     int minDepth(TreeNode *root) {
21         int l = 0, r = 1, mx = 0, mi = 1000000;
22         if(root == NULL) return 0;
23         q[0].p = root;
24         q[0].depth = 1;
25         while(l < r) {
26             if(q[l].p->left == NULL && q[l].p->right == NULL) {
27                 if(q[l].depth < mi) mi = q[l].depth;
28             }
29             if(q[l].p->left != NULL) {
30                 q[r].p = q[l].p->left;
31                 q[r].depth = q[l].depth + 1;
32                 ++r;
33             }
34             if(q[l].p->right != NULL) {
35                 q[r].p = q[l].p->right;
36                 q[r].depth = q[l].depth + 1;
37                 ++r;
38             }
39             ++l;
40         }
41         return mi;
42     }
43 };

Python DFS

1 #111
2 #Runtime: 808 ms (Beats 60.23%)
3 #Memory: 94.9 MB (Beats 26.2%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     min_dep = 1000000
13     def calDepth(self, root, depth):
14         if root.left == None and root.right == None:
15             self.min_dep = min(self.min_dep, depth)
16             return
17         if root.left != None:
18             self.calDepth(root.left, depth + 1)
19         if root.right != None:
20             self.calDepth(root.right, depth + 1)
21         return
22     def minDepth(self, root):
23         """
24         :type root: TreeNode
25         :rtype: int
26         """
27         if root == None:
28             return 0
29         self.calDepth(root, 1)
30         return self.min_dep

Python BFS

1 #111
2 #Runtime: 626 ms (Beats 98.87%)
3 #Memory: 92.3 MB (Beats 79.39%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def minDepth(self, root):
13         """
14         :type root: TreeNode
15         :rtype: int
16         """
17         if not root:
18             return 0
19         q = deque([root])
20         dep = 1
21         while q:
22             sz = len(q)
23             while sz:
24                 sz -= 1
25                 node = q.popleft()
26                 if not node.left and not node.right:
27                     return dep
28                 if node.left:
29                     q.append(node.left)
30                 if node.right:
31                     q.append(node.right)
32             dep += 1
33         return ans

Uriel 2023-07-10 13:58 发表评论

[LeetCode]1601. Maximum Number of Achievable Transfer Requests (Hard) Python3-2023.07.02

Uriel — Sun, 02 Jul 2023 18:10:00 GMT

一共有n个楼�Q�给��Z��些requests�Q�requests_i=[from_i, to_i]�Q�代表一个�h要求从from到to��|��问最多可以满��_��个requests�Q��得最后每栋楼�q�出的�h��C��?br />直接DFS爆搜

1 #1601
2 #Runtime: 1254 ms (Beats 62.13%)
3 #Memory: 16.6 MB (Beats 33.98%)
4
5 class Solution:
6     def maximumRequests(self, n: int, requests: List[List[int]]) -> int:
7         ans = 0
8         remain = [0] * n
9
10         def DFS(x, cnt):
11             nonlocal ans
12             if x == len(requests):
13                 for i in range(n):
14                     if remain[i]:
15                         return
16                 ans = max(ans, cnt)
17                 return
18             remain[requests[x][0]] -= 1
19             remain[requests[x][1]] += 1
20             DFS(x + 1, cnt + 1)
21             remain[requests[x][0]] += 1
22             remain[requests[x][1]] -= 1
23             DFS(x + 1, cnt)
24
25         DFS(0, 0)
26         return ans

Uriel 2023-07-03 02:10 发表评论

[LeetCode]1514. Path with Maximum Probability (Medium) Python3-2023.07.01

Uriel — Sat, 01 Jul 2023 07:30:00 GMT

�l�出n堆不同数量的cookie�Q�问分给k个�h�Q�问拿最多的那个��拿多少个cookie�Q�DFS

1 #1514
2 #Runtime: 66 ms (Beats 76.87%)
3 #Memory: 16.2 MB �Q�Beats 98.88%)
4
5 class Solution:
6     def distributeCookies(self, cookies: List[int], k: int) -> int:
7         cnt = [0] * k
8         ans = float('inf')
9         def DFS(c):
10             nonlocal ans
11             if c == len(cookies):
12                 ans = min(ans, max(cnt))
13                 return
14             for i in range(k):
15                 cnt[i] += cookies[c]
16                 DFS(c + 1)
17                 cnt[i] -= cookies[c]
18                 if not cnt[i]:
19                     break
20
21         DFS(0)
22         return ans

Uriel 2023-07-01 15:30 发表评论

[LeetCode]1970. Last Day Where You Can Still Cross (Hard) Python-2023.06.30

Uriel — Sat, 01 Jul 2023 06:41:00 GMT

一个row行col列的二维矩阵�Q�初始所有元素�ؓ0�Q�代表均可以赎ͼ�每天有一个格子变�?�Q�不可走�Q�问最�q�第几天可以依旧从第一行走到最后一行（具体从第几列开始，走向�W�几列无要求�Q�，可以四个方向�?br />二分�{�案+DFS��认是否可达

1 #1970
2 #Runtime: 3361 ms (Beats 85.71%)
3 #Memory: 39.9 MB (Beats 14.29%)
4
5 class Solution(object):
6     def latestDayToCross(self, row, col, cells):
7         """
8         :type row: int
9         :type col: int
10         :type cells: List[List[int]]
11         :rtype: int
12         """
13         d = [[0, 1], [0, -1], [-1, 0], [1, 0]]
14         def DFS(grid, r, c):
15             if 0 <= r < row and 0 <= c < col and grid[r][c] == 0:
16                 if r == row - 1:
17                     return True
18                 grid[r][c] = -1
19                 for x in d:
20                     tr = r + x[0]
21                     tc = c + x[1]
22                     if DFS(grid, tr, tc):
23                         return True
24             return False
25
26         def ok(x):
27             grid = [[0] * col for _ in range(row)]
28             for i in range(x):
29                 grid[cells[i][0] - 1][cells[i][1] - 1] = 1
30             for i in range(col):
31                 if grid[0][i] == 0 and DFS(grid, 0, i):
32                     return True
33             return False
34
35         l = 1
36         r = len(cells)
37         while l < r:
38             mid = (l + r) // 2 + (l + r) % 2
39             if ok(mid):
40                 l = mid
41             else:
42                 r = mid - 1
43         return l
44

Uriel 2023-07-01 14:41 发表评论

[LeetCode]864. Shortest Path to Get All Keys (Hard) Python-2023.06.29

Uriel — Thu, 29 Jun 2023 14:27:00 GMT

�l�出一个二�l�迷宫，#代表墙，@代表��L��Q�字母a-f代表钥匙�Q�A-F代表对应的锁�Q�如果先拿到了钥匙那么下�ơ经�q�匹配的锁的时候那一格就可以赎ͼ��Q?代表�I�地�Q�问最��多��步可以拿到所有钥匙�?br />BFS�Q�用位操作存储已�l�获得的钥匙情况�Q�且同时记录已经��C��几步

1 #864
2 #Runtime: 305 ms (Beats 78.95%)
3 #Memory: 19.6 MB (Beats 36.84%)
4
5 class Solution(object):
6     def shortestPathAllKeys(self, grid):
7         """
8         :type grid: List[str]
9         :rtype: int
10         """
11         nkey = 0
12         n, m = len(grid), len(grid[0])
13         d = [(0, 1), (1, 0), (0, -1), (-1, 0)]
14         for i in range(n):
15             for j in range(m):
16                 ch = grid[i][j]
17                 if ch == '@':
18                     sx, sy = i, j
19                 if 'a' <= ch <= 'f':
20                     nkey += 1
21         q = deque([(0, sx, sy)])
22         vis = defaultdict(bool)
23         vis[(0, sx, sy)] = True
24         stp = 0
25         while q:
26             sz = len(q)
27             while sz:
28                 sz -= 1
29                 ky, x, y = q.popleft()
30                 if ky == (1 << nkey) - 1:
31                     return stp
32                 for i in d:
33                     tx = x + i[0]
34                     ty = y + i[1]
35                     if 0 <= tx < n and 0 <= ty < m:
36                         tp_ky = ky
37                         ch = grid[tx][ty]
38                         if ch == '#':
39                             continue
40                         if 'a' <= ch <= 'f':
41                             tp_ky |= 1 << (ord(ch) - ord('a'))
42                         if 'A' <= ch <= 'F' and not (ky & (1 << (ord(ch) - ord('A')))):
43                             continue
44                         if not vis[(tp_ky, tx, ty)]:
45                             vis[(tp_ky, tx, ty)] = True
46                             q.append((tp_ky, tx, ty))
47             stp += 1
48         return -1

Uriel 2023-06-29 22:27 发表评论

[LeetCode]1514. Path with Maximum Probability (Medium) Python-2023.06.28

Uriel — Wed, 28 Jun 2023 13:42:00 GMT

�l�出一个无向图�Q�每条边有一个概率��|��问从从start节点到end节点的所有path中概率值最大�ؓ多少�Q�概率��gؓ�q�乘关系�Q?br />

1 #1514
2 #Runtime: 590 ms (Beats 94.20%)
3 #Memory: 26.2 MB (Beats 97.10%)
4
5 class Solution(object):
6     def kSmallestPairs(self, nums1, nums2, k):
7         """
8         :type nums1: List[int]
9         :type nums2: List[int]
10         :type k: int
11         :rtype: List[List[int]]
12         """
13         fg = set()
14         ans = []
15         hp = []
16         for i in range(min(len(nums1), k)):
17             heapq.heappush(hp, (nums1[i] + nums2[0], nums1[i], nums2[0], 0))
18         while k and hp:
19             _, i, j, idx = heapq.heappop(hp)
20             ans.append([i, j])
21             if idx < len(nums2) - 1:
22                 heapq.heappush(hp, (i + nums2[idx + 1], i, nums2[idx + 1], idx + 1))
23             k -= 1
24         return ans

Uriel 2023-06-28 21:42 发表评论

[LeetCode]1161. Maximum Level Sum of a Binary Tree (Medium) Python-2023.06.15

Uriel — Thu, 15 Jun 2023 09:14:00 GMT

求二叉树按层求节点之和，�q�回加和最大的一层的层号�Q�BFS

1 #1161
2 #Runtime: 273 ms (Beats 91.7%)
3 #Memory: 21.5 MB (Beats 67.26%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def maxLevelSum(self, root):
13         """
14         :type root: TreeNode
15         :rtype: int
16         """
17         ans = []
18         q = deque([root])
19         while q:
20             sz = len(q)
21             tp = 0
22             while sz:
23                 sz -= 1
24                 node = q.popleft()
25                 tp += node.val
26                 if node.left:
27                     q.append(node.left)
28                 if node.right:
29                     q.append(node.right)
30             ans.append(tp)
31         return max(range(len(ans)), key=ans.__getitem__) + 1

Uriel 2023-06-15 17:14 发表评论

[LeetCode]530. Minimum Absolute Difference in BST (Easy) Python-2023.06.14

Uriel — Wed, 14 Jun 2023 06:57:00 GMT

求二叉搜索树中元素��g��差最��是多少�Q�DFS中序遍历的同时记录当前节点和上一�ơ访问的节点��g��差即�?br />

1 #530
2 #Runtime: 44 ms (Beats 43.16%)
3 #Memory: 17.3 MB (Beats 71.85%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def getMinimumDifference(self, root):
13         """
14         :type root: TreeNode
15         :rtype: int
16         """
17         self.ans = 100000
18         self.pre_val = None
19
20         def DFS(node):
21             if not node:
22                 return
23             DFS(node.left)
24             if self.pre_val is not None:
25                 self.ans = min(self.ans, node.val - self.pre_val)
26             self.pre_val = node.val
27             DFS(node.right)
28
29         DFS(root)
30         return self.ans

Uriel 2023-06-14 14:57 发表评论

[LeetCode]547. Number of Provinces (Medium) Python-2023.06.04

Uriel — Sun, 04 Jun 2023 15:18:00 GMT

求无向图的联通分支个敎ͼ�DFS | BFS | �q�查�?基本�l�习

DFS

1 #547
2 #Runtime: 154 ms (Beats 66.8%)
3 #Memory: 13.6 MB (Beats 94.53%)
4
5 class Solution(object):
6     def findCircleNum(self, isConnected):
7         """
8         :type isConnected: List[List[int]]
9         :rtype: int
10         """
11         ans = 0
12         n = len(isConnected)
13         vis = [0] * n
14
15         def DFS(node):
16             vis[node] = 1
17             for i in range(n):
18                 if isConnected[node][i] == 1 and not vis[i]:
19                     DFS(i)
20
21         for i in range(n):
22             if not vis[i]:
23                 DFS(i)
24                 ans += 1
25         return ans

BFS

1 #547
2 #Runtime: 156 ms (Beats 61.49%)
3 #Memory: 13.8 MB (Beats 40.4%)
4
5 class Solution(object):
6     def findCircleNum(self, isConnected):
7         """
8         :type isConnected: List[List[int]]
9         :rtype: int
10         """
11         ans = 0
12         n = len(isConnected)
13         vis = [0] * n
14         for i in range(n):
15             if not vis[i]:
16                 q = deque([i])
17                 vis[i] = 1
18                 ans += 1
19                 while q:
20                     x = q.popleft()
21                     for j in range(n):
22                         if not vis[j] and isConnected[x][j]:
23                             vis[j] = 1
24                             q.append(j)
25         return ans

�q�查�?br />

1 #547
2 #Runtime: 178 ms (Beats 21.88%)
3 #Memory: 13.5 MB (Beats 94.53%)
4
5 class Solution(object):
6     def findCircleNum(self, isConnected):
7         """
8         :type isConnected: List[List[int]]
9         :rtype: int
10         """
11         n = len(isConnected)
12         parent = [i for i in range(n)]
13
14         def find(x):
15             if parent[x] != x:
16                 parent[x] = find(parent[x])
17             return parent[x]
18
19         def union(x, y):
20             fa, fb = find(x), find(y)
21             parent[fb] = fa
22
23         for i in range(n):
24             for j in range(n):
25                 if isConnected[i][j]:
26                     union(i, j)
27         return len(set([find(i) for i in range(n)]))

Uriel 2023-06-04 23:18 发表评论

[LeetCode]1376. Time Needed to Inform All Employees (Medium) Python-2023.06.03

Uriel — Sat, 03 Jun 2023 16:27:00 GMT

一个树形结构的公司�Q�所有�h的id�?~n-1�Q�老板的ID为headID�Q�每个�h有个直属上司manager[i]�Q�老板的上�?1�Q�，每个人需要informTime[i]去通知所有下属，问如果老板惌��通知全公司，最快需要多��时��_��DFS

1 #1376
2 #Runtime: 1174 ms (Beats 77.78%)
3 #Memory: 51.1 MB (Beats 38.10%)
4
5 class Solution(object):
6     def numOfMinutes(self, n, headID, manager, informTime):
7         """
8         :type n: int
9         :type headID: int
10         :type manager: List[int]
11         :type informTime: List[int]
12         :rtype: int
13         """
14         son = {}
15         for i in range(n):
16             if manager[i] >= 0:
17                 if manager[i] not in son:
18                     son[manager[i]] = [i]
19                 else:
20                     son[manager[i]].append(i)
21
22         def DFS(id):
23             ans = 0
24             if id not in son:
25                 return 0
26             for i in son[id]:
27                 ans = max(DFS(i), ans)
28             return informTime[id] + ans
29
30         return DFS(headID)

Uriel 2023-06-04 00:27 发表评论

[LeetCode]2101. Detonate the Maximum Bombs (Medium) Python-2023.06.02

Uriel — Fri, 02 Jun 2023 14:58:00 GMT

有一堆二�l�圆形，�l�出每个圆�Ş的中心坐标和半径�Q�如果一个圆形A的覆盖范围包括了另一个圆B的圆心，则选了A��可以cover B�Q�问点击其中某一个最多可以cover几个�?br />
DFS�Q�注意若A能cover B�Q�B不一定能cover A�Q�所以每�ơ要reset vis数组

1 #2101
2 #Runtime: 4068 ms (Beats 9.30%)
3 #Memory: 13.8 MB (Beats 58.14%)
4
5 class Solution(object):
6     def maximumDetonation(self, bombs):
7         """
8         :type bombs: List[List[int]]
9         :rtype: int
10         """
11
12         def In_Range(x, y):
13             if sqrt((bombs[x][1] - bombs[y][1])**2 + (bombs[x][0] - bombs[y][0])**2) <= bombs[x][2]:
14                 return True
15             return False
16
17         def DFS(x):
18             for i in range(len(bombs)):
19                 if not vis[i] and In_Range(x, i):
20                     self.cnt += 1
21                     vis[i] = 1
22                     DFS(i)
23
24         ans = 0
25         for i in range(len(bombs)):
26             vis = [0] * len(bombs)
27             vis[i] = 1
28             self.cnt = 1
29             DFS(i)
30             ans = max(ans, self.cnt)
31         return ans

Uriel 2023-06-02 22:58 发表评论

[LeetCode]1091. Shortest Path in Binary Matrix (Medium) Python-2023.06.01

Uriel — Thu, 01 Jun 2023 15:40:00 GMT

�?-1矩阵从左上角到右下角只走0的话最快几步到达（可以8个方向行赎ͼ��Q�BFS

1 #1091
2 #Runtime: 528 ms (Beats 59.73%)
3 #Memory: 13.8 MB (Beats 71.4%)
4
5 class Solution(object):
6     def shortestPathBinaryMatrix(self, grid):
7         """
8         :type grid: List[List[int]]
9         :rtype: int
10         """
11         if grid[0][0] != 0:
12             return -1
13         if len(grid) == 1 and len(grid[0]) == 1:
14             return 1
15         q = deque([(0, 0, 1)])
16         grid[0][0] = 1
17         d = [[0, 1], [1, 1], [1, 0], [1, -1], [0, -1], [-1, -1], [-1, 0], [-1, 1]]
18         while q:
19             x, y, stp = q.popleft()
20             for i in range(8):
21                 tx = x + d[i][0]
22                 ty = y + d[i][1]
23                 if 0 <= tx < len(grid) and 0<= ty < len(grid[0]) and grid[tx][ty] == 0:
24                     if tx == len(grid) - 1 and ty == len(grid[0]) - 1:
25                         return stp + 1
26                     grid[tx][ty] = 1
27                     q.append((tx, ty, stp + 1))
28         return -1

Uriel 2023-06-01 23:40 发表评论

[LeetCode]934. Shortest Bridge (Medium) Python-2023.05.21

Uriel — Sun, 21 May 2023 14:14:00 GMT

�l�出0-1二维数组�Q?�l�成两个不连通区域，��将几个0改�ؓ1可以让两个区域联�?br />BFS�Q�先从�Q�?开始BFS记录所有boundary cell�Q�然后从boundary cell开始BFS�Q�每�ơ处理完queue中所有元素step+1�Q�知道搜到另一个区�?br />

1 #934
2 #Runtime: 302 ms (Beats 92.21%)
3 #Memory: 13.9 MB (Beats 96.10%)
4
5 class Solution(object):
6     def shortestBridge(self, grid):
7         """
8         :type grid: List[List[int]]
9         :rtype: int
10         """
11         d = [[-1, 0], [1, 0], [0, 1], [0, -1]]
12         x, y = next((x, y) for x in range(len(grid)) for y in range(len(grid[0])) if grid[x][y] == 1)
13         q = deque([(x, y)])
14         boundary = set()
15         while q:
16             x, y = q.popleft()
17             grid[x][y] = -1
18             for i, j in d:
19                 tx = x + i
20                 ty = y + j
21                 if 0 <= tx < len(grid) and 0 <= ty < len(grid[0]) and grid[tx][ty] != -1:
22                     if grid[tx][ty] == 1:
23                         grid[tx][ty] = -1
24                         q.append((tx, ty))
25                     else:
26                         boundary.add((x, y))
27
28         step = 0
29         q = deque(boundary)
30         while q:
31             sz = len(q)
32             while sz > 0:
33                 sz -= 1
34                 x, y = q.popleft()
35                 grid[x][y] = -1
36                 for i, j in d:
37                     tx = x + i
38                     ty = y + j
39                     if 0 <= tx < len(grid) and 0 <= ty < len(grid[0]) and grid[tx][ty] != -1:
40                         if grid[tx][ty] == 1:
41                             return step
42                         else:
43                             grid[tx][ty] = -1
44                             q.append((tx, ty))
45             step += 1
46         return -1
47

Uriel 2023-05-21 22:14 发表评论

[LeetCode]662. Maximum Width of Binary Tree (Medium) Python-2023.04.20

Uriel — Thu, 20 Apr 2023 10:45:00 GMT

判断二叉树所有层中同一层左双��点宽度最大是多少�Q�BFS�Q�每�ơ处理一层�ƈ且求宽度�Q�多开了一个deque来存每个�q�队的节点的�~�号�Q�可以把�~�号和node节点信息合�ƈ�Q�用一个新的node class保存�Q�但��g��用两个deque更省内存

1 #662
2 #Runtime: 28 ms (Beats 72.48%)
3 #Memory: 14.9 MB (Beats 82.57%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def widthOfBinaryTree(self, root):
13         """
14         :type root: TreeNode
15         :rtype: int
16         """
17         q = deque([root])
18         q_idx = deque([1])
19         ans = 0
20         while q:
21             s = len(q)
22             for i in range(s):
23                 node = q.popleft()
24                 x = q_idx.popleft()
25                 if i == 0:
26                     tp_min = x
27                 if i == s - 1:
28                     tp_max = x
29                 if node.left:
30                     q.append(node.left)
31                     q_idx.append(2 * x - 1)
32                 if node.right:
33                     q.append(node.right)
34                     q_idx.append(2 * x)
35             ans = max(ans, tp_max - tp_min + 1)
36         return ans

Uriel 2023-04-20 18:45 发表评论

[LeetCode]1372. Longest ZigZag Path in a Binary Tree (Medium) Python-2023.04.19

Uriel — Wed, 19 Apr 2023 08:08:00 GMT

求二叉树中最长的之字形path�Q�DFS

1 #1372
2 #Runtime: 366 ms (Beats 100%)
3 #Memory: 64.2 MB (Beats 66.67%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def longestZigZag(self, root):
13         """
14         :type root: TreeNode
15         :rtype: int
16         """
17         self.ans = 0
18
19         def DFS(node, pre, length):
20             if not node:
21                 return
22             if length > self.ans:
23                 self.ans = length
24             if pre == -1:
25                 DFS(node.left, 0, length + 1)
26                 DFS(node.right, 1, length + 1)
27             elif pre == 0:
28                 DFS(node.left, 0, 1)
29                 DFS(node.right, 1, length + 1)
30             else:
31                 DFS(node.left, 0, length + 1)
32                 DFS(node.right, 1, 1)
33
34         DFS(root, -1, 0)
35         return self.ans

Uriel 2023-04-19 16:08 发表评论

[LeetCode]1857. Largest Color Value in a Directed Graph (Hard) Python-2023.04.09

Uriel — Sun, 09 Apr 2023 10:28:00 GMT

�l�定一个有向图�Q�和其中每个节点的颜�Ԍ��问其中的所有path中经�q�同色最多的节点的个敎ͼ�DFS划水�q�了
*有部分参�?>https://leetcode.com/problems/largest-color-value-in-a-directed-graph/solutions/3396323/

1 #1857
2 #Runtime: 3256 ms (Beats 12.50%)
3 #Memory: 202.1 MB (Beats 12.50%)
4
5 class Solution(object):
6     def largestPathValue(self, colors, edges):
7         """
8         :type colors: str
9         :type edges: List[List[int]]
10         :rtype: int
11         """
12         graph = defaultdict(list)
13         for x, y in edges:
14             graph[x].append(y)
15         vis = set()
16         path = set()
17         ans = 0
18         n = len(colors)
19         self.has_circle = False
20         cnt = [[0] * 26 for _ in range(n)]
21         def DFS(x):
22             if x in path:
23                 self.has_circle = True
24             if x in vis:
25                 return 0
26             vis.add(x)
27             path.add(x)
28             idx = ord(colors[x]) - ord('a')
29             cnt[x][idx] = 1
30             for y in graph[x]:
31                 DFS(y)
32                 if self.has_circle == True:
33                     return float('inf')
34                 for ch in range(26):
35                     cnt[x][ch] = max(cnt[x][ch], (1 if ch == idx else 0) + cnt[y][ch])
36             path.remove(x)
37             return max(cnt[x])
38
39         for i in range(n):
40             ans = max(ans, DFS(i))
41         return -1 if ans == float('inf') else ans

Uriel 2023-04-09 18:28 发表评论

[LeetCode]133. Clone Graph (Medium) Python-2023.04.08

Uriel — Sat, 08 Apr 2023 10:48:00 GMT

克隆一个无向图�Q�可能有环，直接DFS

1 #133
2 #Runtime: 41 ms (Beats 39.30%)
3 #Memory: 13.8 MB (Beats 37.71%)
4
5 """
6 # Definition for a Node.
7 class Node(object):
8     def __init__(self, val = 0, neighbors = None):
9         self.val = val
10         self.neighbors = neighbors if neighbors is not None else []
11 """
12
13 class Solution(object):
14     def cloneGraph(self, node):
15         """
16         :type node: Node
17         :rtype: Node
18         """
19         def DFS(node, cloned_graph):
20             if node not in cloned_graph:
21                 cloned_graph[node] = Node(node.val)
22                 for nr in node.neighbors:
23                     cloned_graph[node].neighbors.append(DFS(nr, cloned_graph))
24             return cloned_graph[node]
25         if node == None:
26             return None
27         return DFS(node, {})

Uriel 2023-04-08 18:48 发表评论

[LeetCode]1020. Number of Enclaves (Medium) Python-2023.04.07

Uriel — Fri, 07 Apr 2023 11:23:00 GMT

�c�M��1254题，�l�定一�?-1矩阵�Q?代表��岛0代表��_��本题求问有几个格�Ҏ��中的陆地格点（边缘的不��）

同样�Q�DFS求连通分支个敎ͼ�如果不是位于边缘的联通分支，则ans加上此次DFS到达�q�的格点�?br />

1 #1020
2 #Runtime: 691 ms (Beats 37.84%)
3 #Memory: 75.3 MB (Beats 20.27%)
4
5 class Solution(object):
6     def numEnclaves(self, grid):
7         """
8         :type grid: List[List[int]]
9         :rtype: int
10         """
11         self.dir = [[-1, 0], [1, 0], [0, -1], [0, 1]]
12         n = len(grid)
13         m = len(grid[0])
14
15         def DFS(x, y):
16             grid[x][y] = 0
17             self.tp += 1
18             for dx, dy in self.dir:
19                 tx = x + dx
20                 ty = y + dy
21                 if 0 <= tx < n and 0 <= ty < m and grid[tx][ty]:
22                     DFS(tx, ty)
23                 elif tx < 0 or ty < 0 or tx >= n or ty >= m:
24                     self.fg = False
25
26
27         self.vis = [[0] * m for _ in range(n)]
28         self.ans = 0
29         for i in range(1, n - 1):
30             for j in range(1, m - 1):
31                 if grid[i][j] == 1:
32                     self.fg = True
33                     self.tp = 0
34                     DFS(i, j)
35                     if self.fg == True:
36                         self.ans += self.tp
37         return self.ans

Uriel 2023-04-07 19:23 发表评论

[LeetCode]1254. Number of Closed Islands (Medium) Python-2023.04.06

Uriel — Thu, 06 Apr 2023 11:31:00 GMT

�l�定一�?-1矩阵�Q?代表��岛1代表��_��问整个矩阵里有几片陆圎ͼ�边缘的不��）
DFS求连通分支个�?br />

1 #1254
2 #Runtime: 103 ms (Beats 49.4%)
3 #Memory: 14 MB (Beats 37.50%)
4
5 class Solution(object):
6     def closedIsland(self, grid):
7         """
8         :type grid: List[List[int]]
9         :rtype: int
10         """
11         self.dir = [[-1, 0], [1, 0], [0, -1], [0, 1]]
12         n = len(grid)
13         m = len(grid[0])
14
15         def DFS(x, y):
16             grid[x][y] = 1
17             for dx, dy in self.dir:
18                 tx = x + dx
19                 ty = y + dy
20                 if 0 <= tx < n and 0 <= ty < m and not grid[tx][ty]:
21                     DFS(tx, ty)
22                 elif tx < 0 or ty < 0 or tx >= n or ty >= m:
23                     self.fg = False
24
25
26         self.vis = [[0] * m for _ in range(n)]
27         ans = 0
28         for i in range(1, n - 1):
29             for j in range(1, m - 1):
30                 if grid[i][j] == 0:
31                     self.fg = True
32                     DFS(i, j)
33                     if self.fg == True:
34                         ans += 1
35         return ans

Uriel 2023-04-06 19:31 发表评论

[LeetCode]958. Check Completeness of a Binary Tree (Medium) Python-2023.03.15

Uriel — Sat, 01 Apr 2023 08:16:00 GMT

判断一颗二叉树是否是完全的�Q�BFS�Q�如果碰到还有新节点能插入队列，但之前已�l�碰到过NULL节点�Q�表�C�Z��前的某个子树不是完全�?br />

1 #958
2 #Runtime: 22 ms (Beats 58.42%)
3 #Memory: 13.4 MB (Beats 71.73%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 #     def __init__(self, val=0, left=None, right=None):
8 #         self.val = val
9 #         self.left = left
10 #         self.right = right
11 class Solution(object):
12     def isCompleteTree(self, root):
13         """
14         :type root: TreeNode
15         :rtype: bool
16         """
17         gap_between_nodes = False
18         q = deque([root])
19         while q:
20             s = len(q)
21             while s:
22                 s -= 1
23                 node = q.popleft()
24                 if not node:
25                     gap_between_nodes = True
26                 else:
27                     if gap_between_nodes:
28                         return False
29                     q.append(node.left)
30                     q.append(node.right)
31         return True

Uriel 2023-04-01 16:16 发表评论