剑指offer24 — 链表中环的入口结点
题目
给一个链表,若其中包含环,请找出该链表的环的入口结点,否则,输出null。
分析
-
什么是环?
-
判断是否有环的想法1:空间换时间
走一个加一个到set里面,有重复就说明是环。 -
快慢指针:
-
先判断是否有环(判断是否有环的想法2):
快指针走到none,则没环。
快指针和慢指针相遇,则有环。
-
快走2慢走1时,快慢必相遇,相遇时经推导有:
开头至环入口的距离s = 相遇点至环入口距离m + (n-1)*(相遇点至环入口距离m+环入口至相遇点距离d )
-
解
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def EntryNodeOfLoop(self, pHead):
# write code here
# 快慢指针,循环跳
if pHead == None or pHead.next == None:
return None
fastPointer = pHead
slowPointer = pHead
while fastPointer and fastPointer.next:
fastPointer=fastPointer.next.next
slowPointer = slowPointer.next
if fastPointer == slowPointer:
break
fastPointer = pHead
while fastPointer != slowPointer:
fastPointer = fastPointer.next
slowPointer = slowPointer.next
return fastPointer