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142. Linked List Cycle II

时间:2016-09-12 20:24:26      阅读:132      评论:0      收藏:0      [点我收藏+]

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Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space?

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *detectCycle(ListNode *head) {
12         if(head == NULL || head->next == NULL){
13             return NULL;
14         }
15         
16         ListNode* slow = head;
17         ListNode* fast = head;
18         
19         while(fast->next != NULL && fast->next->next != NULL){
20             fast = fast->next->next;
21             slow = slow->next;
22             
23             if(fast == slow){
24                 slow = head;
25                 
26                 while(slow != fast){
27                     slow = slow->next;
28                     fast = fast->next;
29                 }
30                 
31                 return slow;
32             }
33         }
34         return NULL;
35     }
36 };

 

142. Linked List Cycle II

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原文地址:http://www.cnblogs.com/sankexin/p/5866028.html

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