leetcode 728. Self Dividing Numbers

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.

解法1:

养成好习惯，一定要先写测试用例，然后写下伪代码，最后才是代码！切忌直接上代码！

ans = []

for num In range(left, right+1):

　　for each bit in num：

          check num % bit == 0?

      if all bit div is 0 then add num to ans

class Solution(object):
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        [1, 2, 3, 4, 5, 6, 7, 8, 9]
        [11, 12, 15] 10/1,2,5
        [22=2*11, 24=2*2*2*3] 20/2,4,
        [33, 35, 36] 30/3,5,6
        [44, 45, 48] 40/4,5,8
        ...
        [99]
        [111, 112, 115] 110/?
        [122, 124]
        abcd =
        (a*1000+b*100+c*10+d)%a=0
        (a*1000+b*100+c*10+d)%b=0
        (a*1000+b*100+c*10+d)%c=0
        (a*1000+b*100+c*10+d)%d=0
        """
        ans = []
        for i in range(left, right+1):
            if self.is_div_num(i):
                ans.append(i)
        return ans
    
    def is_div_num(self, n):
        if n == 0:
            return False
        q = n
        while q:
            c = q % 10
            if (c == 0) or (n % c != 0):
                return False           
            q /= 10
        return True

改进版本：

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))

关于all：

>>> all([1,2,3])
True
>>> all([1,2,0])
False

官方解法：

class Solution(object):
    def selfDividingNumbers(self, left, right):
        def self_dividing(n):
            for d in str(n):
                if d == '0' or n % int(d) > 0:
                    return False
            return True
        """
        Alternate implementation of self_dividing:
        def self_dividing(n):
            x = n
            while x > 0:
                x, d = divmod(x, 10)
                if d == 0 or n % d > 0:
                    return False
            return True
        """
        ans = []
        for n in range(left, right + 1):
            if self_dividing(n):
                ans.append(n)
        return ans #Equals filter(self_dividing, range(left, right+1))

python2/3 one-liner:
[x for x in range(left, right+1) if all(y and not x%y for y in map(int,str(x)))]

>>> map(int, ["12","23"])
[12, 23]