python打印出1到10的数_Python打印1到最大的n位数

要点是注意输入的n位数是否会导致溢出，因此利用字符串模拟整数的加法。注意：在打印函数中，需要判断打印的数字是否是以0开头的，同时判断条件是 num[i] != "0"，不能写作 num[i] != 0，因为是使用str类型的，后面一种写法导致判断无法成功。'''

输入数字n, 按顺序打印从1最大的n位十进制数

比如输入3, 则打印出1、2、3、到最大的3位数即999

'''

def Print1ToMaxOfNDigits(n):

if n <= 0:

return

number = ['0'] * n

while not Increment(number):

PrintNumber(number)

def Increment(number):

isOverflow = False

nTakeOver = 0

nLength = len(number)

for i in range(nLength-1, -1, -1):

nSum = int(number[i]) + nTakeOver

if i == nLength - 1:

nSum += 1

if nSum >= 10:

if i == 0:

isOverflow = True

else:

nSum -= 10

nTakeOver = 1

number[i] = str(nSum)

else:

number[i] = str(nSum)

break

return isOverflow

def PrintNumber(number):

isBeginning0 = True

nLength = len(number)

for i in range(nLength):

if isBeginning0 and number[i] != '0':

isBeginning0 = False

if not isBeginning0:

print('%c' % number[i], end='')

print('')

#

# Print1ToMaxOfNDigits(2)

def Print1ToMaxOfNDigits2(n):

if n <= 0:

return

number = ['0'] * n

for i in range(10):

number[0] = str(i)

Print1ToMaxOfNDigitsRecursively(number, n, 0)

def Print1ToMaxOfNDigitsRecursively(number, length, index):

if index == length - 1:

PrintNumber(number)

return

for i in range(10):

number[index + 1] = str(i)

Print1ToMaxOfNDigitsRecursively(number, length, index+1)

Print1ToMaxOfNDigits2(2)