求未知长度单链表中倒数第k个节点

https://blog.csdn.net/slibra_L/article/details/78176540

题目：输入一个链表，输出该链表中倒数第k个结点。 
基本思路：遍历一次链表获得链表长度，再次遍历链表，至n-k+1出输出

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if head == None or k <= 0:
            return None
        p = head
        n = 1
        while p.next != None:
            p = p.next
            n = n+1
        if k > n:
            return None
        for i in range(n-k):
            head = head.next
        return head

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进阶思路：设置2个指针，第一个指针走K步之后，第二个指针开始从头走，这样两个指针之间始终相隔K，当指针2走到链表结尾时，指针1的位置即倒数K个节点 
思路推广：寻找中间节点, 两个指针一起, 第一个指针每次走两步, 第二个指针每次走一步, 快指针指到尾部, 慢指针正好指到中间 

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if head == None or k <= 0:
            return None
        p1 = head
        p2 = head
        for i in range(k-1):
            if p1.next == None:
                return None
            p1 = p1.next
        while p1.next != None:
            p1 = p1.next
            p2 = p2.next
        return p2