题目

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。

链接：https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof

思路

快慢指针。快指针先走k步，慢指针再走。当快指针走到最后，慢指针所指的位置就是倒数第k个节点。

特别注意：

  • 链表为空
  • 总结点少于k
  • 输入k为0

代码

  1. # Definition for singly-linked list.
  2. # class ListNode:
  3. #     def __init__(self, x):
  4. #         self.val = x
  5. #         self.next = None
  7. class Solution:
  8.     def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
  9.         # 快慢指针
  10.         fast_ptr = head
  11.         slow_ptr = head
  13.         i = 0
  14.         while fast_ptr is not None:
  15.             if i < k:
  16.                 # fast_ptr = fast_ptr.next
  17.                 i += 1
  18.             else:
  19.                 # fast_ptr = fast_ptr.next
  20.                 slow_ptr = slow_ptr.next
  21.             fast_ptr = fast_ptr.next
  23.         return slow_ptr
  1. # Definition for singly-linked list.
  2. # class ListNode:
  3. #     def __init__(self, x):
  4. #         self.val = x
  5. #         self.next = None
  7. class Solution:
  8.     def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
  9.         # 快慢指针
  10.         # fast_ptr = head
  11.         if head.next is None or k == 0:
  12.             return head
  16.         slow_ptr = head
  18.         i = 0
  19.         while head is not None:
  20.             if i < k:
  21.                 # fast_ptr = fast_ptr.next
  22.                 i += 1
  23.             else:
  24.                 # fast_ptr = fast_ptr.next
  25.                 slow_ptr = slow_ptr.next
  26.             head = head.next
  28.         return slow_ptr if i == k else None