Given a random number generator which can generate the number in range (1,5) uniformly. How can you use it to build a random number generator which can generate the number in range (1,7) uniformly?
(l定一个随机数生成器，q个生成器能均匀生成1?(1,5)的随机数Q如何用这个生成器生成均匀分布??(1,7)的数?)
解法一:
拒绝采样定理
单的? ?1-5 的随机数发生器用两次, 拼成一?q制的数, 是1-25. 这 1-25 q_分配?5U情冉|到7U情况上, 问题p决了. 因ؓ21?的倍数, 我们可以每三个映到一? ?-3 映射?, …, 19-21 映射?. 可见, q些情况之间的概率是一L. 那么, 要是拼成的数字正好是 22-25 q四个呢? 有两U方? W一U是丢弃q个数字, 从头再来, 直到拼成的数字在1-21之间. 因ؓq个是个概率法, 不能保证每次都能落在1-21, 所以采L密度不高. q有一U方? 是说, 假如落到?22-25, 那这ơ的采样l果q上次? 可以证明, q看上去两个互相矛盾的算? l果都能均等的得到等概率的分? (前者叫?Reject Sampling, 后者叫?Metropolis Algorithm, 都是数学物理模拟里面常用的方?
解法?
二进?br />1-2映射?Q?跌Q?-5映射?
生成三位的二q制卛_
simplyzhao 2011-07-19 19:57 发表评论
]]>解法一:
遍历qn个itemsQy妙地利用概率来筛?
void
generate_random_m_from_n(int n, int m)
{
int i, remaining, select = m;
srand(time(NULL));
for(i=0; i<n; i++) {
remaining = n - i;
if(rand()%remaining < select) {
printf("%d\t", i);
--select;
}
}
printf("\n");
}
generate_random_m_from_n(int n, int m)
{
int i, remaining, select = m;
srand(time(NULL));
for(i=0; i<n; i++) {
remaining = n - i;
if(rand()%remaining < select) {
printf("%d\t", i);
--select;
}
}
printf("\n");
}
解法?
shuffleQ即随机z牌E序Q然后选择前m个items卛_
代码参? http://blog.fuqcool.com/2011/04/17/algorithm-shuffle.html
z牌法的一U实?/a>
作者：fuqcool 发布旉Q?011-04-17 23:16:02 分类Q?algorithms
最q自己在做一个小的程序，需要把一个集合里面的元素全部随机地打散。自己想了一个方法，复杂度是nQ觉得不太快。后来参照了一下python关于shuffle的算法，发现我的Ҏ跟它的是一LQ连python的代码都q么写，可能已经没有办法再快了吧Q?/p>
下面来介绍z牌法Q用C语言描述?/p>
法的前提是有一个生随机数的函?/p>
// Generates a random integer between beg and end. int GetRandomNumber(int beg, int end);
q有一个交换函数?/p>
// Swap a and b. void Swap(int a, int b);
上面两个函数我就不写出实CQ因文章的重点在于法的讨论?/p>
假设我们有一堆扑克牌Q怎么才能把这副牌完全打ؕ呢？计算机当然不能像人手那样z牌。但是它可以产生随机敎ͼ随机C一副牌中抽Z张牌是可以的。既然这样那好办了Q我们不停地从牌堆中随机抽取一张扑克牌Q然后把q些牌堆hQ直到原来的牌堆只剩下一张牌的时候ؓ止。这样不完成了z牌的动作了吗?/p>
下面是C代码Q?/p>
int Shuffle(int[] a, int len)
{
for (int i = len - 1; i > 0; i--)
{
// Select an element from index 0 to i randomly;
int index = GetRandomNumber(0, i);
// exchange a[i] with a[index]
Swap(a[index], a[i]);
}
}
Z也脓出python的random单元关于shuffle的实玎ͼ
def shuffle(self, x, random=None, int=int):
"""x, random=random.random -> shuffle list x in place; return None.
Optional arg random is a 0-argument function returning a random
float in [0.0, 1.0); by default, the standard random.random.
"""
if random is None:
random = self.random
for i in reversed(xrange(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = int(random() * (i+1))
x[i], x[j] = x[j], x[i]simplyzhao 2011-07-18 09:32 发表评论
]]>http://zhedahht.blog.163.com/blog/static/25411174200951262930831/
题目Q从扑克牌中随机?/span>5张牌Q判断是不是一个顺子，卌5张牌是不是连l的?/span>2-10为数字本w，A?/span>1Q?/span>J?/span>11Q?/span>Q?/span>12Q?/span>K?/span>13Q而大王可以看成L数字?/span>
思\一:
我们需要把扑克牌的背景抽象成计机语言。不难想象，我们可以?/span>5张牌看成?/span>5个数字组成的数组。大王是特D的数字Q我们不妨把它们都当?/span>0Q这样和其他扑克牌代表的数字׃重复了?/span> 接下来我们来分析怎样判断5个数字是不是q箋的。最直观的是Q我们把数组排序。但值得注意的是Q由?/span>0可以当成L数字Q我们可以用0去补满数l中的空~。也是排序之后的数l不是连l的Q即盔R的两个数字相隔若q个数字Q但如果我们有够的0可以补满q两个数字的I缺Q这个数l实际上q是q箋的。D个例子，数组排序之后?/span>{0Q?/span>1Q?/span>3Q?/span>4Q?/span>5}。在1?/span>3之间I缺了一?/span>2Q刚好我们有一?/span>0Q也是我们可以它当?/span>2d补这个空~?/span>
于是我们需要做三g事情Q把数组排序Q统计数l中0的个敎ͼl计排序之后的数l相L字之间的I缺L。如果空~的L于或者等?/span>0的个敎ͼ那么q个数组是q箋的；反之则不q箋。最后，我们q需要注意的是，如果数组中的?/span>0数字重复出现Q则该数l不是连l的。换成扑克牌的描q方式，是如果一副牌里含有对子，则不可能是顺子?br />
更好的思\?
1Q确?张牌中除?Q其余数字没有重复的Q可以用表统计的ҎQ?
2) 满q样的逻辑Q（maxQmin分别代表5张牌中的?以外的最大值最|
如果没有0Q则max-min=4Q则为顺子，否则不是
如果有一?Q则max-min=4或?Q则为顺子，否则不是
如果有两?Q则max-min=4或?或?Q则为顺子，否则不是
最大值和最值在1Q中可以获得，q样׃用排序了
simplyzhao 2011-07-12 20:33 发表评论
]]>http://zhedahht.blog.163.com/blog/static/25411174200732711051101/
题目Q输入一个正?/span>nQ输出所有和?/span>nq箋正数序列?/span>
例如输入15Q由?/span>1+2+3+4+5=4+5+6=7+8=15Q所以输?/span>3个连l序?/span>1-5?/span>4-6?/span>7-8?/span>
分析Q这是网易的一道面试题?/span>
q道题和本面试题pd的第10?/span>有些cM。我们用两个?/span>small?/span>big分别表示序列的最值和最大倹{首先把small初始化ؓ1Q?/span>big初始化ؓ2。如果从small?/span>big的序列的和大?/span>n的话Q我们向右移?/span>smallQ相当于从序列中L较小的数字。如果从small?/span>big的序列的和小?/span>n的话Q我们向右移?/span>bigQ相当于向序列中dbig的下一个数字。一直到small{于(1+n)/2Q因为序列至要有两个数字?/span>
Zq个思\Q我们可以写出如下代码：
void PrintContinuousSequence(int small, int big);
/////////////////////////////////////////////////////////////////////////
// Find continuous sequence, whose sum is n
/////////////////////////////////////////////////////////////////////////
void FindContinuousSequence(int n)
{
if(n < 3)
return;
int small = 1;
int big = 2;
int middle = (1 + n) / 2;
int sum = small + big;
while(small < middle)
{
// we are lucky and find the sequence
if(sum == n)
PrintContinuousSequence(small, big);
// if the current sum is greater than n,
// move small forward
while(sum > n)
{
sum -= small;
small ++;
// we are lucky and find the sequence
if(sum == n)
PrintContinuousSequence(small, big);
}
// move big forward
big ++;
sum += big;
}
}
/////////////////////////////////////////////////////////////////////////
// Print continuous sequence between small and big
/////////////////////////////////////////////////////////////////////////
void PrintContinuousSequence(int small, int big)
{
for(int i = small; i <= big; ++ i)
printf("%d ", i);
printf("\n");
}
simplyzhao 2011-06-16 19:33 发表评论
]]>#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct Node {
char value;
struct Node *next;
};
struct Node *
list_reverse(struct Node *head)
{
struct Node *tmp, *cur, *pre = NULL;
cur = head;
while(cur) {
tmp = cur->next;
cur->next = pre;
pre = cur;
cur = tmp;
}
return pre;
}
struct Node *
list_reverse_recursive(struct Node *head)
{
struct Node *rv;
if(head && head->next) {
rv = list_reverse_recursive(head->next);
head->next->next = head;
head->next = NULL;
return rv;
} else
return head;
}
void
test_print(struct Node *head)
{
while(head) {
printf("%c\t", head->value);
head = head->next;
}
printf("\n");
}
int
main(int argc, char **argv)
{
struct Node d = {'d', NULL};
struct Node c = {'c', &d};
struct Node b = {'b', &c};
struct Node a = {'a', &b};
test_print(&a);
struct Node *rev_first = list_reverse(&a);
test_print(rev_first);
struct Node *rev_second = list_reverse_recursive(rev_first);
test_print(rev_second);
return 0;
}
#include<stdlib.h>
#include<string.h>
struct Node {
char value;
struct Node *next;
};
struct Node *
list_reverse(struct Node *head)
{
struct Node *tmp, *cur, *pre = NULL;
cur = head;
while(cur) {
tmp = cur->next;
cur->next = pre;
pre = cur;
cur = tmp;
}
return pre;
}
struct Node *
list_reverse_recursive(struct Node *head)
{
struct Node *rv;
if(head && head->next) {
rv = list_reverse_recursive(head->next);
head->next->next = head;
head->next = NULL;
return rv;
} else
return head;
}
void
test_print(struct Node *head)
{
while(head) {
printf("%c\t", head->value);
head = head->next;
}
printf("\n");
}
int
main(int argc, char **argv)
{
struct Node d = {'d', NULL};
struct Node c = {'c', &d};
struct Node b = {'b', &c};
struct Node a = {'a', &b};
test_print(&a);
struct Node *rev_first = list_reverse(&a);
test_print(rev_first);
struct Node *rev_second = list_reverse_recursive(rev_first);
test_print(rev_second);
return 0;
}
simplyzhao 2011-06-12 16:15 发表评论
]]>题目Q有一个复杂链表，其结炚w了有一?font face="Calibri">m_pNext指针指向下一个结点外Q还有一?/span>m_pSibling指向链表中的Ml点或?/span>NULL。其l点?/span>C++定义如下Q?/span>
struct ComplexNode
{
int m_nValue;
ComplexNode* m_pNext;
ComplexNode* m_pSibling;
};
代码Q?
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct Node {
char value;
struct Node *next;
struct Node *random;
};
void test_print(struct Node *);
struct Node *
list_copy_with_random_pointer(struct Node *head)
{
struct Node *tmp, *ptr, *ret;
ptr = head;
while(ptr != NULL) {
tmp = (struct Node *)malloc(sizeof(struct Node));
tmp->value = (ptr->value)-32; /* from lowercase to uppercase, just for testing */
tmp->next = ptr->next;
tmp->random = NULL;
ptr->next = tmp;
ptr = ptr->next->next;
}
ptr = head;
while(ptr != NULL) {
ptr->next->random = ptr->random==NULL ? NULL : ptr->random->next;
ptr = ptr->next->next;
}
ptr = head;
ret = (head==NULL ? NULL : (head->next));
while(ptr != NULL) {
tmp = ptr->next;
ptr->next = ptr->next->next;
tmp->next = ptr->next==NULL ? NULL : ptr->next->next;
ptr = ptr->next;
}
return ret;
}
void
test_print(struct Node *head)
{
while(head != NULL) {
printf("%c: [%c, %c]\n", head->value, head->next==NULL?'-':head->next->value, head->random==NULL?'-':head->random->value);
head = head->next;
}
}
int
main(int argc, char **argv)
{
struct Node d = {'d', NULL, NULL};
struct Node c = {'c', &d, NULL};
struct Node b = {'b', &c, NULL};
struct Node a = {'a', &b, NULL};
a.random = &c;
d.random = &b;
test_print(&a);
struct Node *copy = list_copy_with_random_pointer(&a);
printf("\n\n");
test_print(&a);
printf("\n\n");
test_print(copy);
return 0;
}
#include<stdlib.h>
#include<string.h>
struct Node {
char value;
struct Node *next;
struct Node *random;
};
void test_print(struct Node *);
struct Node *
list_copy_with_random_pointer(struct Node *head)
{
struct Node *tmp, *ptr, *ret;
ptr = head;
while(ptr != NULL) {
tmp = (struct Node *)malloc(sizeof(struct Node));
tmp->value = (ptr->value)-32; /* from lowercase to uppercase, just for testing */
tmp->next = ptr->next;
tmp->random = NULL;
ptr->next = tmp;
ptr = ptr->next->next;
}
ptr = head;
while(ptr != NULL) {
ptr->next->random = ptr->random==NULL ? NULL : ptr->random->next;
ptr = ptr->next->next;
}
ptr = head;
ret = (head==NULL ? NULL : (head->next));
while(ptr != NULL) {
tmp = ptr->next;
ptr->next = ptr->next->next;
tmp->next = ptr->next==NULL ? NULL : ptr->next->next;
ptr = ptr->next;
}
return ret;
}
void
test_print(struct Node *head)
{
while(head != NULL) {
printf("%c: [%c, %c]\n", head->value, head->next==NULL?'-':head->next->value, head->random==NULL?'-':head->random->value);
head = head->next;
}
}
int
main(int argc, char **argv)
{
struct Node d = {'d', NULL, NULL};
struct Node c = {'c', &d, NULL};
struct Node b = {'b', &c, NULL};
struct Node a = {'a', &b, NULL};
a.random = &c;
d.random = &b;
test_print(&a);
struct Node *copy = list_copy_with_random_pointer(&a);
printf("\n\n");
test_print(&a);
printf("\n\n");
test_print(copy);
return 0;
}
simplyzhao 2011-06-09 11:35 发表评论
]]>题目Q定义栈的数据结构，要求d一?/span>min函数Q能够得到栈的最元素。要求函?/span>min?/span>push以及pop的时间复杂度都是O(1)?/span>
#include "StackWithMin.h"
#include<cstdio>
#include<cstdlib>
#include<cstring>
const int StackWithMin::MAX_SIZE ;
StackWithMin::StackWithMin() :
top_index(-1), min_index(-1)
{
printf("StackWithMin Constructor\n");
}
StackWithMin::~StackWithMin()
{
printf("StackWithMin Destructor\n");
}
int
StackWithMin::top() const
{
if(top_index == -1) {
printf("top() failed: Stack Empty\n");
return -1;
}
return stack[top_index];
}
int
StackWithMin::get_min() const
{
if(min_index == -1) {
printf("get_min() failed: Stack Empty\n");
return -1;
}
return stack[min_index];
}
void
StackWithMin::push(int value)
{
if(top_index == -1) { /* stack empty */
stack[++top_index] = value;
min_index = top_index;
return;
}
stack[++top_index] = value;
if(value < stack[min_index]) {
index[top_index] = min_index;
min_index = top_index;
}
}
int
StackWithMin::pop()
{
if(top_index == -1) {
printf("pop() failed: Stack Empty\n");
return -1;
}
int ret = stack[top_index];
if(min_index == top_index)
min_index = index[top_index];
--top_index;
return ret;
}
#include<cstdio>
#include<cstdlib>
#include<cstring>
const int StackWithMin::MAX_SIZE ;
StackWithMin::StackWithMin() :
top_index(-1), min_index(-1)
{
printf("StackWithMin Constructor\n");
}
StackWithMin::~StackWithMin()
{
printf("StackWithMin Destructor\n");
}
int
StackWithMin::top() const
{
if(top_index == -1) {
printf("top() failed: Stack Empty\n");
return -1;
}
return stack[top_index];
}
int
StackWithMin::get_min() const
{
if(min_index == -1) {
printf("get_min() failed: Stack Empty\n");
return -1;
}
return stack[min_index];
}
void
StackWithMin::push(int value)
{
if(top_index == -1) { /* stack empty */
stack[++top_index] = value;
min_index = top_index;
return;
}
stack[++top_index] = value;
if(value < stack[min_index]) {
index[top_index] = min_index;
min_index = top_index;
}
}
int
StackWithMin::pop()
{
if(top_index == -1) {
printf("pop() failed: Stack Empty\n");
return -1;
}
int ret = stack[top_index];
if(min_index == top_index)
min_index = index[top_index];
--top_index;
return ret;
}
class StackWithMin
{
public:
StackWithMin();
~StackWithMin();
int get_min() const;
int top() const;
void push(int value);
int pop();
private:
static const int MAX_SIZE = 101;
int top_index, min_index;
int stack[MAX_SIZE];
int index[MAX_SIZE];
};
simplyzhao 2011-05-25 15:49 发表评论
]]>题目Q输入一二元查找树Q将该二元查找树转换成一个排序的双向链表。要求不能创ZQ何新的结点，只调整指针的指向?
比如二元查找树
10
/ \
6 14
/ \ / \
4 8 12 16
转换成双向链?/span>
4=6=8=10=12=14=16?/span>
思\Q递归Q在一时找不到递归的灵感的时候，多考虑考虑递归的参敎ͼ有时更重要的是考虑递归的返回?br /> 每处理一个节点，首先获取左子树和叛_树所q回的链表，然后拼接
代码Q?
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
/* Problem: Convert a binary search tree into a sorted linkedlist */
/* When it comes to Tree-Structure, recursion is always the most common solution.
When designing recursion solution, should consider:
1. the parameters
2(important). the return object
*/
struct Node {
int value;
struct Node *left;
struct Node *right;
};
struct Node *
BTree2List(struct Node *root)
{
if(root == NULL)
return NULL;
struct Node *ret = NULL;
/* Convert the left tree into a sorted linkedlist */
struct Node *l_linkedlist = BTree2List(root->left);
ret = l_linkedlist==NULL ? root : l_linkedlist;
/* Convert the right tree into a sorted linkedlist */
struct Node *r_linkedlist = BTree2List(root->right);
while(l_linkedlist && l_linkedlist->right)
l_linkedlist = l_linkedlist->right;
/* Combine */
if(l_linkedlist)
l_linkedlist->right = root;
root->left = l_linkedlist;
root->right = r_linkedlist;
if(r_linkedlist)
r_linkedlist->left = root;
return ret;
}
int main(int argc, char** argv)
{
struct Node a = {4, NULL, NULL};
struct Node b = {8, NULL, NULL};
struct Node c = {12, NULL, NULL};
struct Node d = {16, NULL, NULL};
struct Node e = {6, NULL, &b};
struct Node f = {14, &c, NULL};
struct Node g = {10, &e, &f};
struct Node *ret = BTree2List(&g);
while(ret && ret->right) {
ret = ret->right;
}
while(ret) {
printf("%d\n", ret->value);
ret = ret->left;
}
return 0;
}
#include<string.h>
#include<stdlib.h>
/* Problem: Convert a binary search tree into a sorted linkedlist */
/* When it comes to Tree-Structure, recursion is always the most common solution.
When designing recursion solution, should consider:
1. the parameters
2(important). the return object
*/
struct Node {
int value;
struct Node *left;
struct Node *right;
};
struct Node *
BTree2List(struct Node *root)
{
if(root == NULL)
return NULL;
struct Node *ret = NULL;
/* Convert the left tree into a sorted linkedlist */
struct Node *l_linkedlist = BTree2List(root->left);
ret = l_linkedlist==NULL ? root : l_linkedlist;
/* Convert the right tree into a sorted linkedlist */
struct Node *r_linkedlist = BTree2List(root->right);
while(l_linkedlist && l_linkedlist->right)
l_linkedlist = l_linkedlist->right;
/* Combine */
if(l_linkedlist)
l_linkedlist->right = root;
root->left = l_linkedlist;
root->right = r_linkedlist;
if(r_linkedlist)
r_linkedlist->left = root;
return ret;
}
int main(int argc, char** argv)
{
struct Node a = {4, NULL, NULL};
struct Node b = {8, NULL, NULL};
struct Node c = {12, NULL, NULL};
struct Node d = {16, NULL, NULL};
struct Node e = {6, NULL, &b};
struct Node f = {14, &c, NULL};
struct Node g = {10, &e, &f};
struct Node *ret = BTree2List(&g);
while(ret && ret->right) {
ret = ret->right;
}
while(ret) {
printf("%d\n", ret->value);
ret = ret->left;
}
return 0;
}
simplyzhao 2011-05-23 09:09 发表评论
]]>