141. 环形链表
Given a linked list, determine if it has a cycle in it.
To represent a cycle in the given linked list, we use an integer pos
which represents the position (0-indexed) in the linked list where tail connects to. If pos
is -1
, then there is no cycle in the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: true Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0 Output: true Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it using O(1) (i.e. constant) memory?
环形链表。
用快慢指针法:
慢:一次走一步
快:一次走两步
如果快指针走到null则无环
if(f.next==null || f.next.next==null){ return false; }
如果快慢相遇则有环
public class Solution { public boolean hasCycle(ListNode head) { if(head==null||head.next==null||head.next.next==null){ return false; } ListNode s=head; ListNode f=head.next.next; while(s!=f){ if(f.next==null || f.next.next==null){ return false; } s=s.next; f=f.next.next; } return true; } }
用HashSet
把当前处理的链表加入到Set中,如有环,必会找到已存在的链表
api:
add
contains
public class Solution { public boolean hasCycle(ListNode head) { if(head==null||head.next==null||head.next.next==null){ return false; } HashSet<ListNode> hashSet = new HashSet<>(); ListNode cur=head; while (cur!=null){ if(hashSet.contains(cur)){ return true; } hashSet.add(cur); cur=cur.next; } return false; } }