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题意：判断一个链表中是否有环

思路：快慢指针，如果有环，最终快慢指针会在非NULL相遇

注：用到fast->next前先要确保fast非NULL,要用fast->next->next前先要确保fast,fast->next非NULL

复杂度：时间O(n)， 空间O(1)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:


    bool hasCycle(ListNode *head) {
    	ListNode *fast, *slow;
    	fast = slow = head;
    	while(fast && fast->next){
    		fast = fast->next->next;
    		slow = slow->next;
    		if(fast == slow) return true;
    	}
    	return false;
    }