欧美日韩一本到,久久精品国产第一区二区三区最新章节 ,午夜精品久久久久久久蜜桃app

other algorithm

111

posted @ 2006-11-21 22:11 ailab 閱讀(233) | 評論 (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(276) | 評論 (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(232) | 評論 (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(247) | 評論 (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(295) | 評論 (0) | 編輯收藏

dream come true

兩個鏈表是否相交，如果相交，找出第一個開始相交的節點

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(224) | 評論 (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(219) | 評論 (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(228) | 評論 (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(221) | 評論 (0) | 編輯收藏

dream come true!(3)

給定數組a[0:n-1]以及一個正整數k（0<=k<=n-1），設計一個算法，將子數組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(257) | 評論 (0) | 編輯收藏

青青草原综合久久大伊人导航_色综合久久天天综合_日日噜噜夜夜狠狠久久丁香五月_热久久这里只有精品

堅持到底就是勝利

導航

公告

隨筆分類

隨筆檔案

相冊

統計

留言簿(1)

閱讀排行榜

評論排行榜

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)