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# 链表中倒数最后k个结点
 [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/03/pages/R5NyOzkn3qAZy7wCx1pS#简单) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/03/pages/R5NyOzkn3qAZy7wCx1pS#牛客) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/03/pages/R5NyOzkn3qAZy7wCx1pS#双指针)
> [链表中倒数最后k个结点\_牛客题霸\_牛客网](https://www.nowcoder.com/practice/886370fe658f41b498d40fb34ae76ff9)
**问题简述**
```
输入一个长度为 n 的链表，设链表中的元素的值为 ai ，返回该链表中倒数第k个节点。
如果该链表长度小于k，请返回一个长度为 0 的链表。
```
**思路1**
* 先计算链表长度 `L`，再重新走 `L-k` 步；
Python
```python
class Solution:
def FindKthToTail(self , pHead: ListNode, k: int) -> ListNode:
L = 0
cur = pHead
while cur:
cur = cur.next
L += 1
cur = pHead
d = L - k
while d and cur:
cur = cur.next
d -= 1
return cur
```
**思路2：快慢指针**
* 快指针先走 `k` 步；最后返回慢指针；
Python
```python
class Solution:
def FindKthToTail(self , pHead: ListNode, k: int) -> ListNode:
f, s = pHead, pHead
# 快指针先走 k 步
for _ in range(k):
if not f:
return f
f = f.next
while f:
f = f.next
s = s.next
return s
```