首页 > 其他 > 详细

Linked List Cycle II

时间:2014-12-03 13:55:15      阅读:246      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?

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 * 2*x+ind - x = c
 * x+ind = c
 * x = count - ind
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *slow, *fast;
        slow = fast = head;
        int count = 0;
        while(true){
            if(slow==NULL || fast==NULL || slow->next==NULL || fast->next==NULL){
                return NULL;
            }
            slow = slow->next;
            fast = fast->next->next;
            count++;
            if(slow == fast){
                //has a cycle
                while(true){
                    if(head == slow){
                        return slow;
                    }
                    head = head->next;
                    slow = slow->next;
                }
            }
        }
    }
};

 

Linked List Cycle II

原文:http://www.cnblogs.com/code-swan/p/4139667.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!