LeetCode 142. 单链表如何寻找环的入口点（绝对是史上最简单的解法！）

关于带环单链表找入环点的问题，这个问题确实挺难的，网上的大多数解法都看不懂。研究了一下发现了一个更好的解法，其实算法并不是太难，就是得稍微推导一下。以下是推导过程：

package com.xxx;public class MySingleListImpl {    class Node { private int data; private Node next; public Node(int data) {     this.data = data;     this.next = null; } public int getData() {     return this.data; }    }    private Node head;    public MySingleListImpl() { this.head = null;    }//头插法    public void addFirst(int data) { Node node = new Node(data); if (this.head == null) {     this.head = node; }else {     node.next = this.head;     this.head = node; }    }//尾插法    public void addLast(int data) { Node node = new Node(data); if (this.head == null) {     this.head = node; }else {     Node cur = this.head;     while (cur.next != null) {  cur = cur.next;     }     cur.next = node; }    }//打印链表    public void display() { Node cur = this.head; while (cur != null) {     System.out.print(cur.data + " ");     cur = cur.next; } System.out.println();    }    //创建一个有环的链表    public void createLoop() { Node cur = this.head; while (cur.next != null) {     cur = cur.next; } cur.next = this.head.next.next;    }    //寻找环的入口点    public Node detectCycle() { Node fast = this.head; Node slow = this.head; //循环退出两种情况: //1.fast.next == null(找到单链表结尾也没有找到环); 2.fast == slow(有环) while (fast.next != null) {     fast = fast.next.next;     slow = slow.next;     if (fast == slow) { break;     } } //链表没有环的情况 if (fast.next == null) {     return null; } //链表有环 slow = this.head; while (fast != slow) {     fast = fast.next;     slow = slow.next; } return slow;    }}

package com.xxx;public class Test {    public static void main(String[] args) { MySingleListImpl list = new MySingleListImpl(); list.addFirst(134); list.addFirst(20); list.addFirst(35); list.addLast(10); list.addLast(999); list.addLast(7); System.out.print("原链表："); list.display();  list.createLoop(); MySingleListImpl.Node cur = list.detectCycle(); System.out.println("链表入环点是:" + cur.getData());    }}

如果各位大佬有更好的解法，请速速联系我！