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]