题解 | #链表中环的入口结点#

import java.util.*;
/*
 public class ListNode {
    int val;
    ListNode next = null;

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

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        ListNode dummy = new ListNode(-1), slow = dummy, fast = dummy, p = dummy, q = null;
        boolean f = false;
        dummy.next = pHead;

        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;

            if(slow == fast) {
                f = true;
                q = slow;
            }

            if(f) {
                p = p.next;
                q = q.next;
                if(p == q) return p;
            }

        }
        return null;
    }
}

判断链表有无环的升级版。当快慢指针相遇时，设置两个指针p，q；p从头出发，q从slow出发，两指针相遇就是环的入口