题目地址https://leetcode.com/problems/linked-list-cycle/
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?
解法一
/**
* 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) {
ListNode p = head; //游标指针
while (p != null && p.next != null) //注意循环的终止条件!!!!
{
p.val = Integer.MIN_VALUE; //将结点里面的值设为一个超小的值
if (p.next.val == Integer.MIN_VALUE) //判断是否有环
return true;
else
p = p.next;
}
return false;
}
}
解法二
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head, slow = head; //设置快慢指针,fast和slow
boolean flag = false;
while (slow != null && slow.next != null && fast != null && fast.next != null) //要格外注意循环的终止条件!!!!!
{
if (slow == fast && flag == true) //判断两指针是否相遇,且不是在初次出发处
return true;
else
{
flag = true;
slow = slow.next;
fast = fast.next.next;
}
}
return false;
}
}
reference
https://leetcode.com/problems/linked-list-cycle/solution/
题意是给定一个链表,判断此链表里面是否有环,题目的附加要求是只用O(1)的空间解决此题。
Notes
1.链表问题要格外注意循环的终止条件;
LeetCode--LinkedList--141.Linked List Cycle(Easy)
原文:https://www.cnblogs.com/victorxiao/p/11168355.html