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other algorithm

111

posted @ 2006-11-21 22:11 ailab 閱讀(230) | 評(píng)論 (0) | 編輯收藏

tree algorithm

?travel tree without stack or recursive
three travel using none recursive
delete a element in binary search tree
count words(using bst)
level-travel(using queue)

posted @ 2006-11-21 22:11 ailab 閱讀(271) | 評(píng)論 (0) | 編輯收藏

array algorithm

insert a value into sorted array
select nth element
delete duplicate element
find there exits two same items in a array
find duplicate element a[0.N-1]. range from 1--N-1
a+b = s
find prim
occurs odd
findMidArray

posted @ 2006-11-21 22:11 ailab 閱讀(226) | 評(píng)論 (0) | 編輯收藏

list algorithm

reverse_list (none-recursive && recursive)
merge list
find middle element of a list
find the nth element from the end of list
insert a element to a sorted single linked list
sort single linked list
swap every two element in a single linked list
circle && the start point
meet together
ploy multi

?

posted @ 2006-11-21 22:09 ailab 閱讀(242) | 評(píng)論 (0) | 編輯收藏

string algorithm

word_count
reverse_string
reverse_word in a sentence
longest increasing subsequence
longest common subsequence
strstr
trim {space}
compress letter
atoi
atof
find all increasing subsequence
Find out if a string is a palindrome.
Delete characters from string 1 which are present in string 2
?An array of pointers to (very long) strings. Find pointers to the (lexicographically) smallest and largest strings
Given a sequence of characters. How will you convert the lower case characters to upper case characters.

posted @ 2006-11-21 21:58 ailab 閱讀(289) | 評(píng)論 (0) | 編輯收藏

dream come true

兩個(gè)鏈表是否相交，如果相交，找出第一個(gè)開始相交的節(jié)點(diǎn)

node? * list(node? * list1,node? * list2)

{

? if (list1? == ? null ? || ?list2? == ? null )

??? return ? null ;

? int ?len1? = ? 0 ,len2? = ? 0 ;

?node? * p? = ?list1, * q? = ?list2;

? while (p -> next? != ? null )

??? {

????????len1? ++ ;

????????p? = ? -> next;

????????}

?? while (q -> next? != ? null )

?? {

??????len2? ++ ;

??????q? = ?q -> next;

????}

?? if (p? != ?q)

???? return ? null ;

??len1? ++ ;

??len2? ++ ;

??p? = ?list1;q? = ?list2;

?? if (len1? > ?len2?)

?? {

??????? int ?diff? = ?len1? - ?len2;

???????p? = ?list1;

??????? while (diff? > ? 0 ? && ?p != ? null ;)

???????? {

?????????p? = ?p -> next;

?????????diff? -- ;

??????????}

????}

?? else ?

? {

??????? int ?diff? = ?len2? - ?len1;

???????q? = ?list2;

??????? while (diff? > ? 0 ? && ?q? != ? null )

???????? {diff? -- ;?q = ?q -> next;}

??}

? while (p? != ?q)

? {

???p? = ?p -> next;q = ?q -> next;

?}

?? return ?p

????

}

posted @ 2006-11-20 23:32 ailab 閱讀(218) | 評(píng)論 (0) | 編輯收藏

dream come true 5!

reverse list
recursion and none-recursion

node?*reverse_list(node?*head)

{

??if(head?==?null?||?head->next?==?null)

??????return?head;

??node?*cur?=?head->next;

??node?*pre?=?head;

??node?*next?=?null;

??while(cur?!=?null)

??{

?????next?=?cur->next;

?????cur->next?=?pre;

?????pre?=?cur;??

?????cur?=?next;

??}

??head->next?=?null;

??head?=?pre;

??reurn?head;

}

node?*reverse_list(node?*head)

{

??if(head?==?null?||?head->next?==?null)

????return?head;

??node?*cur?=?head->next;

??node?*next?=?null;

??head->next?=?null;

??while(cur?!=?null)

?{

????next?=?cur->next;

????cur->next?=?head;

????head?=?cur;

????cur?=?next;

?}

??return?head;

??

}

posted @ 2006-11-20 22:11 ailab 閱讀(213) | 評(píng)論 (0) | 編輯收藏

dream come true!

node? * merge(node? * head1,node? * head2)

{

?? if (head1? == ? null )

????? return ?head2;

?? if (head2? == ? null )

????? return ?head1;

??reverse_list( & head2);

??node? * head3? = ? null , * cur? = ? null ;

??node? * p? = ?head1, * q? = ?head2;

?? while (p? != ? null ? && ?q? != ? null )

?? {

???? if (p -> value? < ?q -> value)

???? {

??????? if (head3? == ? null )

???????? {

???????????head3? = ?p;

???????????cur? = ?p;

???????????p? = ?p -> next;

?????????}

???????? else

???????? {

??????????cur -> next? = ?p;

??????????cur? = ?p;

??????????p? = ?p -> next;

?????????}

??????}

????? else

?????? {

??????????? if (head3? == ? null )

??????????? {

?????????????head3? = ?q;

?????????????cur? = ?q;

?????????????q? = ?q -> next;

????????????}

??????????? else

????????????? {

???????????????cur -> next? = ?q;

???????????????cur? = ?q;

???????????????q = q -> next;

???????????????}

????????}

???}

??? if (p? == ? null )

??????cur -> next? = ?q;

??? if (q? == ? null )

??????cur -> next? = ?p;

??? return ?head3;

}

posted @ 2006-11-20 22:04 ailab 閱讀(220) | 評(píng)論 (0) | 編輯收藏

dream come true!(4) strstr

there are many implentations

char?*strstr(const?char?*str,const?char?*sub)

{

??if(str?==?null?||?sub?==?null)

????return?null;

??const?char?*p?=?str;

??const?char?*q?=?sub;

??while(*str?!=?'\0'?&&?*sub?!=?'\0')

??{

?????if(*str++?!=?*sub++)

??????{

??????????str?=?++p;

??????????sub?=?q;

????????}

???}

??if(*sub?==?'\0')

????return?p;

??else

????return?null;

}

char?*strstr(const?char?*str,const?char?*sub)

{

??if(str?==?null?||?sub?==?null)

????return?null;

??const?char?*p?=?str;

??const?char?*q?=?sub;

??for(;*str?!=?'\0'?;str++)

??{

????if(*str?!=?*sub)

??????continue;

????p?=?str;

????while(1)

????{

????????if(*sub?==?'\0')

???????????return?str;

????????if(*p++?!=?*sub++)

???????????break;

?????????

??????}?

?????sub?=?q;

}

posted @ 2006-11-20 21:50 ailab 閱讀(215) | 評(píng)論 (0) | 編輯收藏

給定數(shù)組a[0:n-1]以及一個(gè)正整數(shù)k（0<=k<=n-1），設(shè)計(jì)一個(gè)算法，將子數(shù)組a[0:k]和
a[k+1:n-1]交換位置，要求算法在最壞的情況下耗時(shí)O(n)，且用到的輔助空間為O(1)。

in fact,the essence of above question is "revers two times"
(a'b')' = ba;

void?reverse(int?*a,int?n,int?k)

{

??if(a?==?null?||?n?<?0?||?k?>?n)

????return;

??revers_array(a,0,n-1);

??revers_array(a,0,k);

??revers_array(a,k+1,n-1);

??

}

posted @ 2006-11-20 21:20 ailab 閱讀(253) | 評(píng)論 (0) | 編輯收藏

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other algorithm

tree algorithm

array algorithm

list algorithm

string algorithm

dream come true

dream come true 5!

dream come true!

dream come true!(4) strstr

dream come true!(3)