坚持到底就是胜利

用心去做好这件事情

统计

留言簿(1)

阅读排行榜

评论排行榜

2006年11月21日 #

other algorithm

111

posted @ 2006-11-21 22:11 ailab 阅读(143) | 评论 (0)编辑 收藏

tree algorithm

  1.  travel tree without stack or recursive
  2. three travel using none recursive
  3. delete a element in binary search tree
  4. count words(using bst)
  5. level-travel(using queue)

posted @ 2006-11-21 22:11 ailab 阅读(169) | 评论 (0)编辑 收藏

array algorithm

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

posted @ 2006-11-21 22:11 ailab 阅读(113) | 评论 (0)编辑 收藏

list algorithm

  1. reverse_list (none-recursive && recursive)
  2. merge list
  3. find middle element of a list
  4. find the nth element from the end of list
  5. insert a element to a sorted single linked list
  6. sort single linked list
  7. swap every two element in a single linked list
  8. circle && the start point
  9. meet together
  10. ploy multi

 

posted @ 2006-11-21 22:09 ailab 阅读(125) | 评论 (0)编辑 收藏

string algorithm

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

posted @ 2006-11-21 21:58 ailab 阅读(182) | 评论 (0)编辑 收藏

2006年11月20日 #

dream come true

两个链表是否相交,如果相交,找出第一个开始相交的节点

node  * list(node  * list1,node  * list2)
{
 
if (list1  ==   null   ||  list2  ==   null )
   
return   null ;
 
int  len1  =   0 ,len2  =   0 ;
 node 
* =  list1, * =  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 阅读(113) | 评论 (0)编辑 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 阅读(104) | 评论 (0)编辑 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 
* =  head1, * =  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 阅读(110) | 评论 (0)编辑 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 *= str;
  
const char *= 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 *= str;
  
const char *= 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 阅读(125) | 评论 (0)编辑 收藏

dream come true!(3)

给定数组a[0:n-1]以及一个正整数k(0<=k<=n-1),设计一个算法,将子数组a[0:k]和
a[k+1:n-1]交换位置,要求算法在最坏的情况下耗时O(n),且用到的辅助空间为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 阅读(125) | 评论 (0)编辑 收藏

仅列出标题  下一页