输入一个链表，输出该链表中倒数第k个结点

思路：通过反转链表，但是这样的话不好输出单个结点

# -*- 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
        pre = None
        while head:
            first = head.next  ##定义新位置
            head.next = pre    ##将当前位置指向前一个位置
            pre = head
            head = first
        count = 1
        while pre:
            if count == k:
                return pre.val
            pre = pre.next
            count +=1     

利用快慢指针：

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        slow,fast=head,head
        for i in range(k):
            if not fast:
                return None
            fast=fast.next
        while fast:
            slow=slow.next
            fast=fast.next
        return slow