在python中使用递归查找1和最大N位之间的整数

Question

I need to find the integers between 1 and the biggest N-bits integer,eg: when n = 3. I need to return 1...999. And using recursion. The following is my code.The problem is that I don't know the exact data structure to represent the number. Accurate to return (n=2): [1,2,3,....99] , but I return [[0,0],[0,1],..[9,9]] .I use list to represent the number. Does anyone know the exact forn to represent the numbers?

class Solution:
# @param n: An integer.
# return : A list of integer storing 1 to the largest number with n digits.
def setOnebyOne(self,numList,number,n,index):
    if index == n-1:
        print 'index = n-1',n-1,number
        numList.append(number)
        return numList

    print index,'setting',number

    for i in range(10):
        if i == 0:
            number.append(i)
        else:
            number[index+1] = i
        print number
        self.setOnebyOne(numList, number,n,index+1)  

def numbersByRecursion(self, n):
    # write your code here
    if n <1:
        return None
    numList = []
    for i in range(10):
        print i
        number =[]
        print number
        number.append(i)
        print 'number[0]= ',number
        self.setOnebyOne(numList, number,n,0)

Answer 1

This is one way to do it.

class Solution():
    def __init__(self,inp):
        self.inp = inp
        self.val = pow(10,inp) - 1
        self.ans = []

    def solution(self):
        if self.val>0:
            self.ans.append(self.val)
            self.val-=1
            self.solution()


inp = input()
sol = Solution(inp)

sol.solution()
print sol.ans

Also you might wanna see this. Recursion in Python? RuntimeError: maximum recursion depth exceeded while calling a Python object

Python has a recursion depth limit .

Check it out by executing

import sys
print sys.getrecursionlimit()

EDITED

def numbersByRecursion(n,largest,result):`
    def recursion(num,largest,result):
        if num <= largest:
            result.append(num)
            return recursion(num+1,largest,result) 
        else:
            return result
    return recursion(n,largest,result)


result = []
n = input()
largest = pow(10,n) - 1
ans = numbersByRecursion(1,largest,result)
print ans

Answer 2

If your question is how to beat a list like

list = [[0,0], [0,1], [0,2], ... , [9,9]]

into

result_list = [0, 1, 2, 3, ..., 99]

then you could do:

def format_list(list):
    result_list = []
    for item in list:
        result = 0
        power = len(item)*10
        for digit in item:
            result += digit ** power
            power /= 10
        result_list.append(result)
    return result_list

Disclaimer: Didn't test this

Answer 3

Try this

def myfn(n):
    def myfn2(i):
        if i==int(n*'9'):
            return [int(n*'9')]
        return [i]+myfn2(i+1)
    return myfn2(1)

This gives

>>>myfn(2)
[1,2,.....,98,99]

Hope this helps