LeetCode 之 JavaScript 解答第141题 —— 环形链表 I（Linked List Cycle I）

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Time：2019/4/7
Title: Linked List Cycle
Difficulty: Easy

Author:小鹿

题目：Linked List Cycle I

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

Slove:

▉ 算法思路：

两种解题思路：

1）哈希表法：遍历链表，没遍历一个节点就要在哈希表中判断是否存在该结点，如果存在，则为环；否则，将该结点插入到哈希表中继续遍历。

2）用两个快慢指针，快指针走两步，慢指针走一步，如果快指针与慢指针重合了，则检测的当前链表为环；如果当前指针或下一指针为 null ，则链表不为环。

▉ 方法一：哈希表

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

        var hasCycle = function(head) {
            let fast = head;
            let map = new Map();
            while(fast !== null){
                if(map.has(fast)){
                    return true;
                }else{
                    map.set(fast);
                    fast = fast.next;
                }
            }
            return false;
        };

▉ 方法二：快慢指针

 var hasCycle = function(head) {
     if(head == null || head.next == null){
         return false;
     }
     let fast = head.next;
     let slow = head;
     while(slow != fast){
         if(fast == null || fast.next == null){
             return false;
          }
         slow = slow.next;
         fast = fast.next.next;
     }
     return true;
 };

▉ 方法二：快慢指针

这部分代码是我自己写的，和上边的快慢指针思路相同，运行结果相同，但是当运行在 leetcode 时，就会提示超出时间限制，仔细对比代码，我们可以发现，在逻辑顺序上还是存在差别的，之所以超出时间限制，是因为代码的运行耗时长。

//超出时间限制
var hasCycle = function(head) {
    if(head == null || head.next == null){
        return false;
    }
    let fast = head.next;
    let slow = head;
    while(fast !== null && fast.next !== null){
        if(slow === fast) return true;
        slow = head.next;
        fast = fast.next.next;
    }
    return false;
};

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