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

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題目：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. 復(fù)制代碼

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. 復(fù)制代碼

Example 3:

Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list. 復(fù)制代碼

Follow up:

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

Slove:

▉ 算法思路：

兩種解題思路：

1）哈希表法：遍歷鏈表，沒(méi)遍歷一個(gè)節(jié)點(diǎn)就要在哈希表中判斷是否存在該結(jié)點(diǎn)，如果存在，則為環(huán)；否則，將該結(jié)點(diǎn)插入到哈希表中繼續(xù)遍歷。

2）用兩個(gè)快慢指針，快指針走兩步，慢指針走一步，如果快指針與慢指針重合了，則檢測(cè)的當(dāng)前鏈表為環(huán)；如果當(dāng)前指針或下一指針為 null ，則鏈表不為環(huán)。

▉ 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;};復(fù)制代碼

▉ 方法二：快慢指針

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;}; 復(fù)制代碼

▉ 方法二：快慢指針

這部分代碼是我自己寫(xiě)的，和上邊的快慢指針?biāo)悸废嗤?#xff0c;運(yùn)行結(jié)果相同，但是當(dāng)運(yùn)行在 leetcode 時(shí)，就會(huì)提示超出時(shí)間限制，仔細(xì)對(duì)比代碼，我們可以發(fā)現(xiàn)，在邏輯順序上還是存在差別的，之所以超出時(shí)間限制，是因?yàn)榇a的運(yùn)行耗時(shí)長(zhǎng)。

//超出時(shí)間限制 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; }; 復(fù)制代碼

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