打印 1 到最大的 n 位数（C++ 和 Python 实现）

（说明：本博客中的题目、题目详细说明及参考代码均摘自 “何海涛《剑指Offer：名企面试官精讲典型编程题》2012年”）

题目

　　输入数字 n，按顺序打印出从 1 到最大的 n 位十进制数．比如输入 3，则打印出 1,2,3 一直到最大的 3 位数即 999 .

算法设计思想

　　由于最大的 n 位十进制可能超过整型范围的限制，而成为大数问题．本题目的关键是如何实现大数的表示或运算．本博客采用参考书中的两种方法，将从 1 到最大 n 位数之间的所有数都看作 n 位数，实际的数若不足 n 位，则在前补 0．具体的设计思想如下 ：
　　1) 使用字符串模拟数字加法，从 1 开始递增到最大 n 位数．
　　在计算机中，n 位数可用包含 n 个指定字符（ '0' - '9' ）的字符串（所有字符均为 '0' 除外）表示．其可以想象为对字符串实现伪码： 　

for ( i = ; i < max_n_digits; i++ ) print i;

　　2) 将 n 位数看做是 n 个数（0 - 9）的排列问题．
　　n 位数的排列问题，即 n 位数的每一位都可取 10 个数（0 - 9）中任意一个数，对于 n 位数，共有 10^n 个选择，注意需要去掉所有位都是 0 的排列，与上一个方法的输出结果相同．

易错点：在打印每个数时，打印前导零是没有意义的．其中，前导零是第一个非零元素的最高有效位之前的所有零．

C++实现

/*
* Author: klchang
* Date: 2018.2.26
* Description: Print digits from 1 to the maximum n digits.
*/
#include <iostream>
#include <string>

// Check if the string contains illegal characters
bool checkDigitString(std::string numeric_str)
{
    bool isLegal = true;

    std::basic_string<char>::iterator iter = numeric_str.begin();
    for (; iter != numeric_str.end(); ++ iter) {
        char ch = *iter;
        if (ch < '' || ch > '') {
            isLegal = false;
            break;
        }
    }
    return isLegal;
}

// Remove the leading zeros in a numeric string
std::string removeLeadingZeros(std::string numeric_str)
{
    int i = ;
    size_t len = numeric_str.length();

    // Return null string when including illegal characters
    if (!checkDigitString(numeric_str)) {
        std::cout << "Input string " << numeric_str << " contains at least an illegal character." << std::endl;
        return "";
    }

    for (; i < len; ++i) {
        if (!(numeric_str[i] == ''))
            break;
    }
    if (i >= len) {
        numeric_str = "";
    } else {
        numeric_str = numeric_str.substr(i);
    }

    return numeric_str;
}

// Simulate the numeric operation that numeric string adds one
std::string incrementByOne(std::string& numeric_str)
{
    size_t len = numeric_str.size();
    std::string output_str(numeric_str);

    if (len <= )
        return output_str;

    int carry = ;
    bool lowest_bit = true;
    std::basic_string<char>::reverse_iterator riter = output_str.rbegin();
    for (; riter != output_str.rend(); ++ riter) {
        int value = *riter - '';
        if (lowest_bit) {
            lowest_bit = false;
            value ++;
        }
        value += carry;
        carry = ;  // clear carry
        if (value > ) {
            carry = ;
            value -= ;
        }
        *riter = '' + value;  // update correspondent characters
        if (carry <= )  break;
    }
    // pass the length of number string
    if (carry > ) {
        output_str = std::string("") + output_str;
    }

    return output_str;
}

// Compare the two numeric strings
// Return value: int,
//               1 when s1 > s2; 0 when s1 == s2; -1 when s1 < s2
int compare(std::string s1, std::string s2)
{
    int result = ;
    std::string valid_s1, valid_s2;
    valid_s1 = removeLeadingZeros(s1);
    valid_s2 = removeLeadingZeros(s2);
    size_t len_1 = valid_s1.size();
    size_t len_2 = valid_s2.size();

    if (len_1 > len_2) {
        result = ;
    } else if (len_1 < len_2) {
        result = -;
    } else {
        std::basic_string<char>::iterator iter1 = valid_s1.begin();
        std::basic_string<char>::iterator iter2 = valid_s2.begin();
        for (; iter1 != valid_s1.end(); ++iter1, ++iter2 ) {
            if (*iter1 == *iter2) {
                continue;
            } else if (*iter1 > *iter2) {
                result = ;
            } else {
                result = -;
            }
            break;
        }
    }

    return result;
}

// Print the digits without the leading zeros
void printDigits(std::string digits)
{
    std::string out_str = removeLeadingZeros(digits);
    if (out_str != "") {
        std::cout << out_str << std::endl;
    }
}

// Print N digits of length `length` starting from index start
void printNDigitsRecursively(char* digits, int length, int start)
{
    if (length == start) {
        std::string cur_number = digits;
        printDigits(cur_number);
        return;
    }
    // Set the digit of the start index
    for (int i = ; i < ; ++ i) {
        digits[start] = i + '';
        printNDigitsRecursively(digits, length, start+);
    }
}

// print digits from 1 to maximum
void printNDigits(int n, int method=)
{
    if (n <= ) {
        std::cout << "ERROR: Illegal parameters n <= 0!" << std::endl;
        return;
    }

    if (method == ) {
        // Recursive method
        std::cout << "\nUse the recursive method to print the numbers from 1 to maximum n digits: " << std::endl;
        int start = ;
        char* digits = new char[n+];
        digits[n] = '\0';  // easy to forget to add the '\0' to the C string

        for (int i = ; i < ; ++i) {
            digits[] = i + '';  // Convert digit to character digits
            printNDigitsRecursively(digits, n, start);
        }

        delete[] digits;
    } else {
        // Simulation of the integer self-incrementation operation method
        // for (i = 0; i < 10; ++i) print i
        std::cout << "\nSimulate the self incrementation of the integer: " << std::endl;
        // Construct maximum integer in string form
        std::string maxNdigits("");
        for (int i = ; i < n; ++i)
            maxNdigits += "";

        std::string cur_num = "";
        do {
            std::cout << cur_num << std::endl;
            cur_num = incrementByOne(cur_num);
        } while (compare(maxNdigits, cur_num) >= );

        std::cout << std::endl;
    }
}

void unitest()
{
    int n = ;
    printNDigits(n, );  // recursive method
    printNDigits(n, );  // simulation method
}

int main()
{
    unitest();

    return ;
}

Python 实现

#!/usr/bin/python
#-*- coding: utf8 -*-
"""
# Author: klchang
# Date: 2018.2.26
# Description: Print digits from 1 to the maximum n digits.
"""

# Generic interface to print numbers from 1 to the maximum of n digits
def print_n_digits(n, method=0):
    if n <= 0:
        print("ERROR: Illegal parameters n <= 0!")
        return
    if method == 1:
        print("\nUse the recursive method to print the numbers from 1 to maximum n digits: ")
        string = ["" for i in range(n)]
        print_n_digits_recursive(string, n, 0)
    else:
        print("\nSimulate the self incrementation of the integer: ")
        print_n_digits_simulation(n)

# Use the recursive method to print
def print_n_digits_recursive(string, length, start):
    if length == start:
        output = remove_leading_zeros("".join(string))
        if output != '':
            print(output)
        return

    for i in range(0,10):
        string[start] = repr(i)
        print_n_digits_recursive(string, length, start+1)

# Use the simulation method to print
def print_n_digits_simulation(n):
    max_num = '' * n
    curr_num = ''
    while True:
        print(curr_num)
        next_num = increment_by_one(curr_num)  # add by 1 function: i += 1
        # Check if next_num > max_num
        if compare(next_num, max_num) > 0:  # compare function: i > , <, or == some_number
            break
        else:
            curr_num = next_num

def remove_leading_zeros(num):
    i = 0
    for ch in num[:-1]:
        if '' == ch:
            i += 1
        else:
            break
    return num[i:]   

# Compare two digits string
def compare(num_1, num_2):
    result = 0

    real_num_1 = remove_leading_zeros(num_1)
    real_num_2 = remove_leading_zeros(num_2)
    len_1 = len(real_num_1)
    len_2 = len(real_num_2)

    if len_1 == len_2:
        if real_num_1 > real_num_2:
            result = 1
        elif real_num_1 < real_num_2:
            result = -1
        else:
            result = 0
    else:
        if len_1 > len_2:
            result = 1
        elif len_1 < len_2:
            result = -1
        else:
            result = 0

    return result

def check_digits_string(num):
    is_legal = True

    # Check if the input num string is legal or not
    # Check the type
    if not isinstance(num, str):
        is_legal = False
        print("Input param is not str type!")
    # Check the length
    num_length = len(num)
    if num_length <= 0:
        is_legal = False
        print('Illegal String: null str')
    # Check the characters contained
    legal_chars = set([repr(i) for i in range(10)])
    for ch in num:
        if ch not in legal_chars:
            print("Input number string includes non-digit character")
            is_legal = False
            break

    return is_legal

def increment_by_one(num):
    next_ = ''
    num = remove_leading_zeros(num)
    # First, check that it is a legal input number string.
    if not check_digits_string(num):
        return next_
    # The effective length of num
    num_length = len(num)
    # The Least Significant Bit
    carry = False
    char_code = ord(num[-1]) + 1
    out_seq, index = [], 0
    for i in range(num_length-2, -1, -1):
        if char_code > ord(''):
            carry = True
            out_seq.append('')
        else:
            out_seq.append(chr(char_code))
            index = i + 1
            break
        # Process the case with carry
        char_code = ord(num[i]) + 1
        carry = False

    # Reverse the output sequence
    out_seq.reverse()
    next_ = ''.join(out_seq)
    if char_code > ord(''):
        next_ = '' + next_    # Overflow
    elif index == 0:
        next_ = chr(char_code) + next_
    else:
        next_ = num[:index] + next_

    return next_

def unitest():
    n = 3
    print_n_digits(n, 1)  # recursive method
    print_n_digits(n, 0)  # simulation method

if __name__ == '__main__':
    unitest()

参考代码

1. targetver.h

#pragma once

// The following macros define the minimum required platform.  The minimum required platform
// is the earliest version of Windows, Internet Explorer etc. that has the necessary features to run
// your application.  The macros work by enabling all features available on platform versions up to and
// including the version specified.

// Modify the following defines if you have to target a platform prior to the ones specified below.
// Refer to MSDN for the latest info on corresponding values for different platforms.
#ifndef _WIN32_WINNT            // Specifies that the minimum required platform is Windows Vista.
#define _WIN32_WINNT 0x0600     // Change this to the appropriate value to target other versions of Windows.
#endif

2. stdafx.h

// stdafx.h : include file for standard system include files,
// or project specific include files that are used frequently, but
// are changed infrequently
//

#pragma once

#include "targetver.h"

#include <stdio.h>
#include <tchar.h>

// TODO: reference additional headers your program requires here

3. stdafx.cpp

// stdafx.cpp : source file that includes just the standard includes
// Print1ToMaxOfNDigits.pch will be the pre-compiled header
// stdafx.obj will contain the pre-compiled type information

#include "stdafx.h"

// TODO: reference any additional headers you need in STDAFX.H
// and not in this file

4. Print1ToMaxOfNDigits.cpp

// Print1ToMaxOfNDigits.cpp : Defines the entry point for the console application.
//

// 《剑指Offer――名企面试官精讲典型编程题》代码
// 著作权所有者：何海涛

#include "stdafx.h"
#include <memory>

void PrintNumber(char* number);
bool Increment(char* number);
void Print1ToMaxOfNDigitsRecursively(char* number, int length, int index);

// ====================方法一====================
void Print1ToMaxOfNDigits_1(int n)
{
    if(n <= )
        return;

    char *number = new char[n + ];
    memset(number, '', n);
    number[n] = '\0';

    while(!Increment(number))
    {
        PrintNumber(number);
    }

    delete []number;
}

// 字符串number表示一个数字，在 number上增加1
// 如果做加法溢出，则返回true；否则为false
bool Increment(char* number)
{
    bool isOverflow = false;
    int nTakeOver = ;
    int nLength = strlen(number);

    for(int i = nLength - ; i >= ; i --)
    {
        int nSum = number[i] - '' + nTakeOver;
        if(i == nLength - )
            nSum ++;

        if(nSum >= )
        {
            if(i == )
                isOverflow = true;
            else
            {
                nSum -= ;
                nTakeOver = ;
                number[i] = '' + nSum;
            }
        }
        else
        {
            number[i] = '' + nSum;
            break;
        }
    }

    return isOverflow;
}

// ====================方法二====================
void Print1ToMaxOfNDigits_2(int n)
{
    if(n <= )
        return;

    char* number = new char[n + ];
    number[n] = '\0';

    for(int i = ; i < ; ++i)
    {
        number[] = i + '';
        Print1ToMaxOfNDigitsRecursively(number, n, );
    }

    delete[] number;
}

void Print1ToMaxOfNDigitsRecursively(char* number, int length, int index)
{
    if(index == length - )
    {
        PrintNumber(number);
        return;
    }

    for(int i = ; i < ; ++i)
    {
        number[index + ] = i + '';
        Print1ToMaxOfNDigitsRecursively(number, length, index + );
    }
}

// ====================公共函数====================
// 字符串number表示一个数字，数字有若干个0开头
// 打印出这个数字，并忽略开头的0
void PrintNumber(char* number)
{
    bool isBeginning0 = true;
    int nLength = strlen(number);

    for(int i = ; i < nLength; ++ i)
    {
        if(isBeginning0 && number[i] != '')
            isBeginning0 = false;

        if(!isBeginning0)
        {
            printf("%c", number[i]);
        }
    }

    printf("\t");
}

// ====================测试代码====================
void Test(int n)
{
    printf("Test for %d begins:\n", n);

    Print1ToMaxOfNDigits_1(n);
    Print1ToMaxOfNDigits_2(n);

    printf("Test for %d ends.\n", n);
}

int _tmain(int argc, _TCHAR* argv[])
{
    Test();
    Test();
    Test();
    Test();
    Test(-);

    return ;
}

5. 参考代码下载

项目 12_Print1ToMaxOfNDigits 下载： 百度网盘

何海涛《剑指Offer：名企面试官精讲典型编程题》 所有参考代码下载：百度网盘

参考资料

[1] 何海涛. 剑指 Offer：名企面试官精讲典型编程题 [M]. 北京：电子工业出版社，2012. 94-99.