题解 | #链表中环的入口结点#

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        slow = pHead
        fast = pHead
        while slow:
            slow = slow.next
            # 判断环
            if fast.next:
                fast = fast.next.next
                if not fast:
                    return None
                # 若有环，则从相遇点到环入口的步数=从起点到环入口的步数
                if fast == slow:
                     p = pHead
                     q = fast
                     while p and q:
                        if p == q:
                            return p
                        p = p.next
                        q = q.next
                    # 只需要遍历2次链表，所以是O(n)
        return None