题解 | #链表中环的入口结点#

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        if (pHead==null || pHead.next==null) return null;
        ListNode fast = pHead;
        ListNode slow = pHead;
        while (fast!=null && fast.next!=null) {
            fast=fast.next.next;
            slow=slow.next;
            // 快慢相遇，说明成环
            if (fast==slow) {
                ListNode tmp = pHead;
                while (tmp!=slow) {
                    tmp=tmp.next;
                    slow=slow.next;
                }
                return tmp;
            }
        }
        return null;
    }
}

解题思想：

* 方式一：借助set

* 方式二：快慢指针