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?

题意：判断链表是否有环，若有环返回环起始的结点。

思路：设计两个快慢指针，快的走两步，慢的走一步，则如果相遇则说明有环，这个时候把快指针从头开始走，一次走一步，再次与慢指针相遇的时候，则就是环的起点（证明根据快指针走的距离是慢指针距离的两倍）。若无环，则返回NULL。

/** * 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) {ListNode *fast = head, *slow = head;while (fast && fast->next){fast = fast->next->next;slow = slow->next;if (fast == slow){fast = head;while (fast != slow){slow = slow->next;fast = fast->next;}return fast;}}return NULL;}};