# 链表中倒数第k个节点
**1.** **Link**
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**2. 剑指 Offer 22. 链表中倒数第k个节点**
难度简单188收藏分享切换为英文接收动态反馈
输入一个链表，输出该链表中倒数第k个节点。为了符合大多数人的习惯，本题从1开始计数，即链表的尾节点是倒数第1个节点。
例如，一个链表有 `6` 个节点，从头节点开始，它们的值依次是 `1、2、3、4、5、6`。这个链表的倒数第 `3` 个节点是值为 `4` 的节点。
**示例：**
```
给定一个链表: 1->2->3->4->5, 和 k = 2.
返回链表 4->5.
```
**3. 思路**
1. **方法1：**
1. **先把list 过一遍找到list length**
2. **计算正着数的node的position，即 list length - k**
3. **再从头遍历一遍找到对应位置的node，并返回**
2. **方法2**
1. **把list直接先reverse**
2. **找到正着数第k个node**
3. **把head 达到这个node的list再reverse一遍并返回**
3. **两种方法都是Time complexity: O(n), Space complexity: O(1)因为只是单纯把linked list 线性遍历，没有用到额外的空间**
**4. Coding**
Method **1**: 正找node的位置
```
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
#
#1. method 1:
# loop through whole list and get the length of list
# then find the position of the kth node from tail
#
#2. method2:
# first reverse list
# then find the kth element from head
# then reverse the sublist to get the Kth From end
#
# Two methods have time O(n) and space O(1)
#base case:
#
if not head:
return None
cur = head
length = 0
while cur:
cur = cur.next
length +=1
if k > length:
return None
pos = length - k
cur_pos = 0
cur = head
while cur_pos!=pos:
cur = cur.next
cur_pos += 1
return cur
```
Method 2: reverse list
```
class Solution:
def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
# method 2
#
if not head:
return None
cur = new_head = self.reverse(head)
pos = 1
while pos != k:
cur = cur.next
pos += 1
cur.next = None
res = self.reverse(new_head)
return res
def reverse(self, node):
prev, cur = None, node
while cur:
next_node = cur.next
cur.next = prev
prev = cur
cur = next_node
return prev
```
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