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Linked List Cycle II

时间:2014-12-03 13:57:58      阅读:163      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.

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

标签:style   blog   io   color   sp   for   on   div   log   

原文地址:http://www.cnblogs.com/code-swan/p/4139667.html

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