0141._Linked_List_Cycle

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0141. Linked List Cycle

难度: Easy

刷题内容

原题连接

内容描述

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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:

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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.

img

Example 2:

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

img

Example 3:

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

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Follow up:

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

解题方案

思路 1
**- 时间复杂度: O(N)**- 空间复杂度: O(N)**

使用快慢指针的思路进行解题。就像两个运动员在同一个环形赛道上赛跑，如果一个运动员跑的快，一个跑得慢，最后两个运动员一定会相遇。

下面代码中的fast每次会走两步，而slow每次会走一步，如果fast没有next节点，自然没有环；如果fast等于slow说明二者相遇，最终为表明存在环。

执行结果

执行用时 :92 ms, 在所有 JavaScript 提交中击败了94.16%的用户

内存消耗 :36.6 MB, 在所有 JavaScript 提交中击败了51.93%

代码：

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