【算法详解】打印1到最大的n位数

1. 问题描述

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.

注意的问题：考虑大数的情形导致的溢出。

2. 算法1

使用字符串数组表示大数，在字符数组上模拟整数的加法(加 1 ); 

字符串数组中每个元素都是 ‘0’ 到 ‘9’ 之间的某一个字符。初始化时将每一个字符都设置为 ‘0’， 相当于整数 0 .

#include <iostream>
#include <string>

using namespace std;

// print the integers characters array.
// if current interger is 123, the integer characters array looks like as below
// a[0], a[1], a[2], a[3]
// '3'   '2'   '1'   '0'
// the order is opposite to real interger order, so we need to print the characters array form the first non '0' element with reversed order.
void printInteger(char* pStrIntegers, int size)
{
    if (pStrIntegers == NULL || size <= 0)
    {
        return;
    }
    
    bool isPrint = false;
    
    for (int i = size - 1; i >= 0; i--)
    {
        if (!isPrint)
        {
            if(pStrIntegers[i] != '0')
            {
                isPrint = true;
                cout<<pStrIntegers[i];
            }
        }else
        {
            cout<<pStrIntegers[i];
        }
    }
    
    cout<<endl;
}

// add 1 to current integer, mock addition operation on the charaters array.
bool incrementone(char* pStrIntegers, int size)
{
    if (pStrIntegers == NULL || size <= 0)
    {
        return false;
    }
    
    bool isValid = true;
    
    // for 0 ~ size-1 th elements
    // try to add 1 from the first character: pStrIntegers[0]
    // from 1th element to nth elements
    for (int i = 0; i < (size - 1); i++)
    {
        if (pStrIntegers[i] != '9')
        {
            pStrIntegers[i] = pStrIntegers[i] + 1;  // if current character is not '9', then just add 1 on current characte.
            break;                                  // end this increment, add 1 successfully.
        }else
        {
            // current character is '9', adding 1 should carry.
            // go to next character element, meanwhile set current character to '0'.carry
            pStrIntegers[i] = '0';
            
            // if current character is in the last valid index, this increment is invalid, as current character should be like "999..999".
            if (i == (size - 2))
            {
                isValid = false;
            }
        }
    }
    
    return isValid;
}

void printToMaxOfInteger(int n)
{
    if (n <= 0)
    {
        cout<<"Invalid n = "<<n<<endl;
        return;
    }
    
    cout<<"Start to print n = "<<n<<endl;
    
    // declare a character array with capacity n + 1
    char strIntegers[n + 1];
    // initialize all the characters as '0'.
    for (int i = 0; i <= n; i++)
    {
        strIntegers[i] = '0';
    }
    
    // start to increment by 1.
    while (incrementone(strIntegers, n + 1))
    {
        printInteger(strIntegers, n + 1);
    }
}


int main(int argc, const char * argv[])
{
    printToMaxOfInteger(3);
    printToMaxOfInteger(-1);
    printToMaxOfInteger(0);
    
    return 0;
}

2. 算法2

使用全排列。

参考：剑指offer 名企面试官精讲典型编程题》 - 面试题12：打印1到最大的n位数