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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; } }
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