【1错1对0】链表环入口 Linked List Cycle I

https://www.nowcoder.com/practice/253d2c59ec3e4bc68da16833f79a38e4?tpId=13&tqId=11208&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
and
https://leetcode.com/problems/linked-list-cycle-ii/

题目：判断单链表是否有环，并输出环入口
思路：

1. 快慢指针是否相遇判断有环与否；
2. 两指针同速，分别从起点和相遇点开始遍历，再次相遇的点即为环入口

public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        if(pHead == null || pHead.next == null) {
            return null;
        }
        
        ListNode pSlow = pHead;
        ListNode pFast = pHead;
        
        while(pFast != null && pFast.next != null) {
            pSlow = pSlow.next;
            pFast = pFast.next.next;
            
            if(pSlow == pFast) {
                pSlow = pHead;
                while(pSlow != pFast) {
                    pSlow = pSlow.next;
                    pFast = pFast.next;
                }
                return pSlow;
            }
        }
        return null;
        
        
    }
}