解題思路:
分別設置2個指針(s,q)指向鏈表的頭部,s每次指向下面一個(s = s.next),q每次指向下面2個(q = q.next.next).
如果存在環,q總會在某一時刻追上s

/*** Definition for singly-linked list.* function ListNode(val) {*     this.val = val;*     this.next = null;* }*//*** @param {ListNode} head* @return {boolean}*/
var hasCycle = function(head) {let s = head;let f = head;while(f && f.next){s = s.next;f = f.next.next;if(s === f) return true}return false;
};