Python 从 0 到 n 的数字中出现的数字的次数

发布于09月23日

给定一个数字n，计算包括n在内的数字0、2和4的出现次数.

示例1:

n = 10
output: 4

Example2:

n = 22
output: 11

我的代码:

n = 22

def count_digit(n):
    count = 0
    for i in range(n+1):
        if '2' in str(i):
            count += 1
        if '0' in str(i):
            count += 1
        if '4' in str(i):
            count += 1
    return count

count_digit(n)

编码输出:10

==同步，由长者更正==

约束:1 <= N <= 10^5

注:The solution should not cause 100 or 101 for large numbers.

推荐答案

另一种 brute 的力量，似乎更快:

def count_digit(n):
    s = str(list(range(n+1)))
    return sum(map(s.count, '024'))

基准:n = 10**5:

result   time   solution

115474  244 ms  original
138895   51 ms  Kelly
138895  225 ms  islam_abdelmoumen
138895  356 ms  CodingDaveS

代码(Try it online!):

from timeit import default_timer as time

def original(n):
    count = 0
    for i in range(n+1):
        if '2' in str(i):
            count += 1
        if '0' in str(i):
            count += 1
        if '4' in str(i):
            count += 1
    return count

def Kelly(n):
    s = str(list(range(n+1)))
    return sum(map(s.count, '024'))

def islam_abdelmoumen(n):
    count = 0
    for i in map(str,range(n+1)):
        count+=i.count('0')
        count+=i.count('2')
        count+=i.count('3')
    return count

def CodingDaveS(n):
    count = 0
    for i in range(n + 1):
        if '2' in str(i):
            count += str(i).count('2')
        if '0' in str(i):
            count += str(i).count('0')
        if '4' in str(i):
            count += str(i).count('4')
    return count

funcs = original, Kelly, islam_abdelmoumen, CodingDaveS

print('result   time   solution')
print()
for _ in range(3):
    for f in funcs:
        t = time()
        print(f(10**5), ' %3d ms ' % ((time()-t)*1e3), f.__name__)
    print()