[LeetCode] 142.环形链表II

1. 问题描述

142. 环形链表 II - 力扣（LeetCode）

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2. 实现思路

2.1 自己的思路

~~（一定要写出来啊喂_）~

• 受到上一个题哈希表的启发，这个题也可以用哈希表？

时间与空间复杂度都是O(n)

/*
 * @lc app=leetcode.cn id=142 lang=csharp
 *
 * [142] 环形链表 II
 */

var s = new Solution();
s.DetectCycle();

public class ListNode
{
    public int val;
    public ListNode next;
    public ListNode(int x)
    {
        val = x;
        next = null;
    }
}

// @lc code=start
public class Solution
{
    public ListNode DetectCycle(ListNode head)
    {
        if (head == null) return null;
        var h = head;
        ISet<ListNode> visited = new HashSet<ListNode>();
        while (h.next != null)
        {
            if (visited.Contains(h)) return h;
            visited.Add(h);
            h = h.next;
        }
        return null;
    }
}
// @lc code=end

2.2 题解思路

142.环形链表II | 快慢指针 | 环入口查找

• 判断是否有环：快慢指针相遇
• 判断环的入口：Floyd判圈算法

public class Solution
{
    public ListNode DetectCycle(ListNode head)
    {
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
            if (fast == slow)
            {
                fast = head;
                while (fast != slow)
                {
                    fast = fast.next;
                    slow = slow.next;
                }
                return fast;
            }
        }
        return null;
    }
}

2.3 优化（或者敲一遍题解？）

/*
 * @lc app=leetcode.cn id=142 lang=csharp
 *
 * [142] 环形链表 II
 */

var s = new Solution();
s.DetectCycle();

public class ListNode
{
    public int val;
    public ListNode next;
    public ListNode(int x)
    {
        val = x;
        next = null;
    }
}

// @lc code=start
public class Solution
{
    public ListNode DetectCycle(ListNode head)
    {
        if (head == null) return null;
        var slow = head;
        var fast = head;
        while (fast != null && fast.next != null)
        {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast)
            {
                fast = head;
                while (fast != slow)
                {
                    fast = fast.next;
                    slow = slow.next;
                }
                return fast;
            }
        }
        return null;
    }
}
// @lc code=end

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3. 补充知识と杂谈，总结