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[LeetCode] Linked List Cycle II

时间:2014-11-04 10:38:41      阅读:150      评论:0      收藏:0      [点我收藏+]

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

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

 

思路:快慢指针的应用。时间复杂度O(n),空间复杂度O(1)

 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) return head;
13         ListNode *fast = head;
14         ListNode *slow = head;
15         
16         while (fast != NULL && fast->next != NULL) {
17             fast = fast->next->next;
18             slow = slow->next;
19             if (fast == slow) break;
20         }
21         
22         if (fast == NULL || fast->next == NULL) 
23             return NULL;
24             
25         slow = head;
26         while (slow != fast) {
27             slow = slow->next;
28             fast = fast->next;
29         }
30         return slow;
31     }
32 };

 

[LeetCode] Linked List Cycle II

原文:http://www.cnblogs.com/vincently/p/4072867.html

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