链表中倒数最后k个结点:快慢指针fast先走k步,然后fast/slow同步走;注意处理空值情形
链表中倒数最后k个结点
https://www.nowcoder.com/practice/886370fe658f41b498d40fb34ae76ff9
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
#
# @param pHead ListNode类
# @param k int整型
# @return ListNode类
#
class Solution:
def FindKthToTail(self , pHead: ListNode, k: int) -> ListNode:
# write code here
# 快慢指针,fast先移动k步,然后slow、fast同时移动,直至fast为空
# 此时slow位于倒数第k个位置,return slow
fast=pHead
slow=pHead
for i in range(k):
if not fast:
return None # k步还没走完 fast就成了空值,则返回空值
fast=fast.next
while fast:
slow=slow.next
fast=fast.next
return slow
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