Leetcode 142. Linked List Cycle II

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?

s思路：
1. 如何用快慢指针大法检测cycle，在Leetcode 141. Linked List Cycle以及说清楚了。不说了，直接来！

//方法1：快慢指针移动大法！class Solution {public:    ListNode *detectCycle(ListNode *head) {        //        if(!head) return NULL;        ListNode* fast=head->next,*slow=head;        while(fast&&fast!=slow){            fast=fast->next?fast->next->next:NULL;            slow=slow->next;        }        if(fast==NULL) return NULL;        fast=head;        slow=slow->next;//这里有一个bug：把fast放在头部，那么slow需要往下移动一位，才开始同步移动！        while(fast!=slow){            fast=fast->next;            slow=slow->next;        }        return fast;    }};