Given a linked list, determine if it has a cycle in it.
给定一个链表,判断链表中是否有环。
关键技巧:快慢指针,如果有环的话,快指针一定会追上慢指针。
/**
* 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 || head.next == null){
return false;
}
ListNode quick = head;
ListNode slow = head;
while (quick.next != null && quick.next.next != null){
quick = quick.next.next;
slow = slow.next;
if (quick == slow){
return true;
}
}
return false;
}
}