LeetCode--LinkedList--141.Linked List Cycle（Easy）-优快云博客

博客围绕LeetCode 141题“Linked List Cycle”展开，题目要求判断给定链表是否有环，且只用O(1)空间。介绍了两种解法，一是遍历链表修改节点值判断，二是利用快慢指针有环必相遇思想，同时提醒链表问题要注意循环终止条件。

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141. Linked List Cycle（Easy）2019.7.10

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?

solution

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/

总结

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。

解法一是直接遍历链表，将链表里面当前结点的val值设置为一个极小的数Integer.MIN_VALUE，然后判断当前结点的下一个结点值val是否为Integer.MIN_VALUE,若是，则返回true，否则再判断下一个结点，如果循环结束还没找到，则返回false。
解法二利用了快慢指针有环必相遇的思想。首先设置一个slow指针，每次走一步，然后设置一个fast指针，每次走两步，若链表有环，则若干步后两个指针必相遇。解法里面设置了一个flag变量来控制第一次slow==fast时不会返回true。此种解法要格外注意while循环的结束条件，稍不留意，就会出错。

Notes
1.链表问题要格外注意循环的终止条件；

转载于:https://www.cnblogs.com/victorxiao/p/11168355.html