剑指 Offer 22. 链表中倒数第k个节点 (Python 实现)

x33g5p2x  于2022-06-29 转载在 Python  
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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。

示例:

给定一个链表: 1->2->3->4->5, 和 k = 2.
返回链表 4->5.

code:

  1. # Definition for singly-linked list.
  2. # class ListNode:
  3. # def __init__(self, x):
  4. # self.val = x
  5. # self.next = None
  6. class Solution:
  7. def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
  8. temp = head
  9. n = 0
  10. while temp:
  11. n+=1
  12. temp = temp.next
  13. for _ in range(n-k):
  14. head = head.next
  15. return head
  1. # Definition for singly-linked list.
  2. # class ListNode:
  3. # def __init__(self, x):
  4. # self.val = x
  5. # self.next = None
  6. class Solution:
  7. def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
  8. fast, slow = head, head
  9. for _ in range(k-1):
  10. fast = fast.next
  11. while fast.next != None:
  12. fast = fast.next
  13. slow = slow.next
  14. return slow

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