Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
一刷
题解：
方法1， 用hashMap保存节点是否被visited的信息。如果新的节点已经存在了hashmap中，则有环。
方法2, 快慢指针，快指针每次前进两步，慢指针每次前进一步。如果慢指针和快指针重逢，说明有环。
/**
* 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 slow = head;
ListNode fast = head.next;
while(fast!=null && fast.next!=null){
if(fast == slow) return true;
fast = fast.next.next;
slow = slow.next;
}
return false;
}
}
二刷
快慢指针
/**
* 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 slow = head, fast = head.next;
while(fast!=null && fast.next!=null){
if(fast == slow) return true;
fast = fast.next.next;
slow = slow.next;
}
return false;
}
}
