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. 如何用快慢指針大法檢測cycle，在Leetcode 141. Linked List Cycle以及說清楚了。不說了，直接來！
//方法1：快慢指針移動大法！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;//這里有一個bug：把fast放在頭部，那么slow需要往下移動一位，才開始同步移動！ while(fast!=slow){ fast=fast->next; slow=slow->next; } return fast; }};新聞熱點
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