Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
判断一个链表中是否有环存在,用快慢指针来解决,设定两个指针fast和slow都指向head,我们让fast = fast.next.next,slow = slow.next,如果存在环,fast和slow就会相遇,因此当fast == slow时我们就返回true, 否则返回false。代码如下:
Follow up:
Can you solve it without using extra space?
判断一个链表中是否有环存在,用快慢指针来解决,设定两个指针fast和slow都指向head,我们让fast = fast.next.next,slow = slow.next,如果存在环,fast和slow就会相遇,因此当fast == slow时我们就返回true, 否则返回false。代码如下:
/**
* 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) {
if(head == null || head.next == null) return false;
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if(slow == fast)
return true;
}
return false;
}
}