Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
判断是否成环
1、利用set,很简单,但是题目中说不要用额外的空间。
/**
* 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) { Set<ListNode> set = new HashSet<ListNode>(); while( head != null ){
if( set.contains(head) )
return true;
set.add(head);
head = head.next;
}
return false; }
}
2、设置两个指针,一个每次走两个,一个每次走一个,如果走到头,那么返回false,如果相同,那么返回true。
/**
* 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.next;
ListNode slow = head; while( fast.next != null && fast.next.next != null ){
slow = slow.next;
fast = fast.next.next;
if( fast == slow )
return true;
} return false; }
}