首页 > 其他 > 详细

Linked List Cycle II

时间:2014-05-30 16:00:38      阅读:374      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?

bubuko.com,布布扣
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) 
    {
        if(head==NULL || head->next==NULL) return NULL;
        //check if cycle exist
        ListNode* p1=head;
        ListNode* p2=head;
        while(true)
        {
            p1=p1->next;
            p2=p2->next;
            if(p2==NULL || p2->next==NULL) return NULL;
            p2=p2->next;
            if(p1==p2) break;
        }
        p2=head;
        while(p1!=p2)
        {
            p1=p1->next;
            p2=p2->next;
        }
        return p1;
    }
};
bubuko.com,布布扣

 

Linked List Cycle II,布布扣,bubuko.com

Linked List Cycle II

原文:http://www.cnblogs.com/erictanghu/p/3759713.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!