剑指offer-题15：链表中倒数第k个节点

题目描述

输入一个链表，输出该链表中倒数第k个结点。

实验平台：牛客网

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解决思路：

另外，我们还需要注意以下会出现的问题，来增强代码的鲁棒性：

1.输入的链表的头结点为null

2.链表的节点总数少于k个

3.输入的参数k为0

针对问题1和3，我们可以返回null;针对问题2，我们可以在循环时进行判断即可，代码如下：

java:

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        if (head == null || k == 0) {
            return null;
        }
        ListNode firstNode = head;
        ListNode secondeNode = null;

        for (int i = 0; i < k - 1; i++) {
            if (firstNode.next != null) {
                firstNode = firstNode.next;
            }else {
                return null;
            }
        }

        secondeNode = head;
        while (firstNode.next != null) {
            firstNode = firstNode.next;
            secondeNode = secondeNode.next;
        }
        return secondeNode;
    }
}

python:

# -*- 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 is None or k == 0:
            return None
        first_node = head
        for i in range(0, k-1):
            if first_node.next is not  None:
                first_node = first_node.next
            else:
                return None
        second_node = head
        while first_node.next is not None:
            first_node = first_node.next
            second_node = second_node.next
        return second_node