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;
      
    }
}