【LeetCode】728. Self Dividing Numbers 解題報告（Python）

做者： 負雪明燭
id： fuxuemingzhu
我的博客： http://fuxuemingzhu.cn/

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目錄

• • 題目描述
  • 題目大意
  • 解題方法
  • • 循環
    • filter函數
    • 數字迭代
  • 日期

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

題目描述

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.

題目大意

若是一個數字能被它本身的各位數字整除，那麼這個數字是一個自除數字，求在[left, right]雙閉區間內的全部自除數字。

解題方法

循環

用了兩個函數，一個用來判斷是不是dividing number，另外一個用來循環和遍歷。

要注意的一點是要判斷0是否在num中，不然有除0錯誤。

dividing number 判斷的有點麻煩，就是遍歷每位數字。

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函數

參考了https://leetcode.com/problems/self-dividing-numbers/discuss/109445。
有更簡單的兩個函數：
all()判斷是否是全部的元素都知足，
filter過濾掉不知足條件的元素。

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))

數字迭代

轉成字符串的方法耗時，其實能夠直接使用數字求餘的方法節省了大量的時間。

時間複雜度是O(N)，空間複雜度是O(1)。戰勝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，有點累