【面試題22】鏈表中倒數第k個節點

Python題解

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def FindKthToTail(self, head, k):
        if not head or k == 0:
            return None
        ahead = head
        behind = head
        for i in range(k-1):
            if ahead.next != None:
                ahead = ahead.next
            else:
                return None
        while ahead.next != None:
            ahead = ahead.next
            behind = behind.next
        return behind

考點

• 考查對鏈表的理解；
• 考查縮寫代碼的魯棒性。魯棒性是解決這道題的關鍵所在，如果寫出的代碼有多處崩潰的潛在風險，那麼很難通過面試。