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;
}
}
原文：https://www.cnblogs.com/iwyc/p/15202290.html