链表的倒数第k个节点

微信图片_20191221162931.jpg

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if head is None:
            return None
        first = head
        second = head
        for _ in range(k):
            if first == None:
                return None
            first = first.next
        while first != None:
            first = first.next
            second = second.next
        return second