160. Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

 

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

 

Notes:

  • If the two linked lists have no intersection at all, return null.
  • The linked lists must retain their original structure after the function returns.
  • You may assume there are no cycles anywhere in the entire linked structure.
  • Your code should preferably run in O(n) time and use only O(1) memory.
    找到相交的结点，很简单的题目，用双指针记录两个链表，如果相等则返回结点，如果没有则一直next下去，尽头没有next了就返回空。当其中一个指针走完的时候，就重定向到另外一个链表，这样子就可以让两个指针步数一致。在第二次迭代同时碰到交叉点结点。最后的时间复杂度是两个链表的长度之和，考虑最坏情况O(m+n)。例:A={1,2,3,4,5},B={6,7,8,3,4,5}。A的长度比B小，A先走完了，此时B走到4的位置，A重定向后在链表B的6的位置，此时B走到5，然后B重定向到A链表的位置1，最后两个指针距离交叉点3的步数一致。
     1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
 9     if(!headA || !headB)    return NULL;
10     struct ListNode *pa=headA,*pb=headB;
11     while(1){
12         if(pa==pb)  
13             return pa;
14         else if(!pa->next && !pb->next) 
15             return NULL;
16         else{
17             if(pa->next)    
18                 pa=pa->next;
19             else    
20                 pa=headB;
21             if(pb->next)
22                 pb=pb->next;
23             else
24                 pb=headA;
25         }
26     } 
27 }

     

    看到别的大神更简洁的写法,同样的原理:
     1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
 9     if(!headA || !headB)    return NULL;
10     struct ListNode *pa=headA,*pb=headB;
11     while(pa!=pb){
12         pa=pa==NULL?headB:pa->next;
13         pb=pb==NULL?headA:pb->next;
14     }
15     return pa;
16 }

     

posted @ 2018-11-01 00:22  杜王町上班族  阅读(169)  评论(0)  编辑  收藏  举报