Problem: 141. 环形链表

题目描述

思路

定义快慢指针fast、slow，当fast != null && fast.next != null时fast每次走两步、slow走一步，当fast和slow相遇时，则说明存在环

复杂度

时间复杂度:

$O(n)$

空间复杂度:

$O(1)$

Code

/*** Definition for singly-linked list.* class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
class Solution {/*** Linked List Cycle** @param head The head of linked list* @return boolean*/public boolean hasCycle(ListNode head) {if (head == null || head.next == null) {return false;}ListNode fast = head;ListNode slow = head;while (fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;if (fast == slow) {return true;}}return false;}
}