标签:详解 efi link rmi == turn you false with
Given a linked list, determine if it has a cycle in it.
给定一个链表,确定它是否有一个循环。
Follow up:
Can you solve it without using extra space?
你能不用额外的空间解决它吗?
1 /** 2 * Definition for singly-linked list. 3 * class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public boolean hasCycle(ListNode head) { 14 ListNode slow=head,fast=head; 15 while(fast!=null && fast.next!=null){ 16 slow=slow.next; 17 fast=fast.next.next; 18 if(fast==slow) return true; 19 } 20 return false; 21 } 22 }
快慢指针。
设两个指针,慢指针一次走一步,快指针一次走两步,链表有环,必相遇
标签:详解 efi link rmi == turn you false with
原文地址:https://www.cnblogs.com/chanaichao/p/9608813.html