[Force] button 141. The circular linked list

First, the subject description:

Given a list, the list is determined whether a ring.

To show the list given ring, we integer pos connected to the end of the list to represent the position of the list (index starts from 0). If pos is -1, the ring is not on this list.

Example 1:

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

Example 2:

输入：head = [1,2], pos = 0
输出：true
解释：链表中有一个环，其尾部连接到第一个节点。

Example 3:

输入：head = [1], pos = -1
输出：false
解释：链表中没有环。

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/linked-list-cycle
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

Second, the subject of analysis:

Using speed pointers, slow down a step, moving twice fast, slow and fast coincidence indicates a cycloalkyl.

Third, Code Description:

/**
 * 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) {
    //slow和fast都指向head
        ListNode slow=head;
        ListNode fast=head;
       while(fast!=null&&fast.next!=null){
           slow=slow.next;
           fast=fast.next.next;
           //两个节点重合，说明链表带环。
           if(slow==fast){
               return true;
           }
       }
       return false;
    }
}