3.链表中环的入口节点

题目描述

对于一个给定的链表，返回环的入口节点，如果没有环，返回null。

代码实现：

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if (head == null || head.next == null) {
            return null;
        }
        
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                fast = head;
                while (fast != slow) {
                    slow = slow.next;
                    fast = fast.next;
                }
                return fast;
            } 
        }
        return null;
    }
}