title: Linked List Cycle
tags:
- linked-list-cycle
- No.141
- simple
- list
- pointer
Description
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
Corner Cases
- empty list
Solutions
Fast and Slow Pointers
Set a fast pointer and a slow one. When the slow meets the fast, then a cycle exists.
/**
* 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) {
if (head == 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;
}
}
