【LeetCode】728. Self Dividing Numbers 解题报告（Python）

做者： 负雪明烛
id： fuxuemingzhu
我的博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • 循环
    • filter函数
    • 数字迭代
  • 日期

题目地址：https://leetcode.com/problems/self-dividing-numbers/description/

题目描述

A self-dividing number is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

Example 1:

Input: 
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

Note:

The boundaries of each input argument are 1 <= left <= right <= 10000.

题目大意

若是一个数字能被它本身的各位数字整除，那么这个数字是一个自除数字，求在[left, right]双闭区间内的全部自除数字。

解题方法

循环

用了两个函数，一个用来判断是不是dividing number，另外一个用来循环和遍历。

要注意的一点是要判断0是否在num中，不然有除0错误。

dividing number 判断的有点麻烦，就是遍历每位数字。

class Solution:
    def isDividingNumber(self, num):
        if '0' in str(num):
            return False
        return 0 == sum(num % int(i) for i in str(num))
    def selfDividingNumbers(self, left, right):
        """ :type left: int :type right: int :rtype: List[int] """
        answer = []
        for num in range(left, right+1):
            print(num)
            if self.isDividingNumber(num):
                answer.append(num)
        return answer

filter函数

参考了https://leetcode.com/problems/self-dividing-numbers/discuss/109445。
有更简单的两个函数：
all()判断是否是全部的元素都知足，
filter过滤掉不知足条件的元素。

class Solution(object):
    def selfDividingNumbers(self, left, right):
        is_self_dividing = lambda num: '0' not in str(num) and all([num % int(digit) == 0 for digit in str(num)])
        return filter(is_self_dividing, range(left, right + 1))

As pointed out by @ManuelP, [num % int(digit) == 0 for digit in str(num)] creates an entire list which is not necessary. By leaving out the [ and ], we can make use of generators which are lazy and allows for short-circuit evaluation, i.e. all will terminate as soon as one of the digits fail the check.

The answer below improves the run time from 128 ms to 95 ms:

class Solution(object):
    def selfDividingNumbers(self, left, right):
        is_self_dividing = lambda num: '0' not in str(num) and all(num % int(digit) == 0 for digit in str(num))
        return filter(is_self_dividing, range(left, right + 1))

数字迭代

转成字符串的方法耗时，其实能够直接使用数字求余的方法节省了大量的时间。

时间复杂度是O(N)，空间复杂度是O(1)。战胜98%.

class Solution:
    def selfDividingNumbers(self, left, right):
        """ :type left: int :type right: int :rtype: List[int] """
        res = []
        for num in range(left, right + 1):
            if self.isDividing(num):
                res.append(num)
        return res
        
    def isDividing(self, num):
        temp = num
        while temp:
            div = temp % 10
            if not div or num % div != 0:
                return False
            temp //= 10
        return True

日期

2018 年 1 月 13 日 2018 年 11 月 5 日 —— 打了羽毛球，有点累