LeetCode实战：环形链表 II_tail connects to node index 7 预期结果 tail connects t

作者：神奇cpp | 2024-08-20 04:21:12

踩

tail connects to node index 7 预期结果 tail connects to node index 9

背景

题目英文

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

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.

Note: Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
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Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
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Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
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Follow-up:

Can you solve it without using extra space?

题目中文

给定一个链表，返回链表开始入环的第一个节点。如果链表无环，则返回 null。

为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。如果 pos 是 -1，则在该链表中没有环。

说明：不允许修改给定的链表。

示例 1：

输入：head = [3,2,0,-4], pos = 1
输出：tail connects to node index 1
解释：链表中有一个环，其尾部连接到第二个节点。
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示例 2：

输入：head = [1,2], pos = 0
输出：tail connects to node index 0
解释：链表中有一个环，其尾部连接到第一个节点。
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示例 3：

输入：head = [1], pos = -1
输出：no cycle
解释：链表中没有环。
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进阶：

你是否可以不用额外空间解决此题？

算法实现

利用暴力匹配的方式

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode DetectCycle(ListNode head)
    {
        if (head == null)
            return null;

        ListNode p1 = head;
        int i = 0;
        while (p1 != null)
        {
            p1 = p1.next;
            i++;

            ListNode p2 = head;
            int j = 0;
            while (j < i)
            {
                if (p2 == p1)
                {
                    return p2;
                }
                p2 = p2.next;
                j++;
            }
        }
        return null;
    }
}
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利用Hash的方式

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode DetectCycle(ListNode head)
    {
        if (head == null)
            return null;

        HashSet<ListNode> hash = new HashSet<ListNode>();
        hash.Add(head);

        ListNode temp = head.next;
        while (temp != null)
        {
            if (hash.Contains(temp))
                return temp;

            hash.Add(temp);
            temp = temp.next;
        }
        return null;
    }
}
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实验结果