标签：https xpl return tno 两个指针 inpu ant nec OLE
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?
快慢指针的应用(慢的一次一下，快的一次两下）
/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public boolean hasCycle(ListNode head) { ListNode slow = head; ListNode fast = head; while(fast!=null && fast.next != null){ fast = fast.next; if(slow == fast) return true; slow = slow.next; fast = fast.next; } return false; } }
// Linked List Cycle // 时间复杂度O(n)，空间复杂度O(1) class Solution { public boolean hasCycle(ListNode head) { // 设置两个指针，一个快一个慢 ListNode slow = head, fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; } return false; } };
标签：https xpl return tno 两个指针 inpu ant nec OLE
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/10541410.html
