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