Description:

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

My code:

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    if(!head) {
        return false;
    }
    if(!head.next) {
        return false;
    }
    let slow = head, fast = head.next;
    while(fast) {
        if(fast == slow) {
            return true;
        } else {
            fast = fast.next;
            if(!fast) {
                return false;
            } else {
                fast = fast.next;
                slow = slow.next;
            }
        }
    }
    return false;
};

Note: 兩個指針，一個步長大，一個步長小，如果在某個時刻相等則表明有環(huán)。步長大的是分步進(jìn)行的，在第一次next時還要判斷是否為null，如果是就返回false，否則再兩個指針next，不然可能會出現(xiàn)null不存在next的情況。