剑指offer-面试题12：打印1到最大的n位数

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.

解法：这道提的陷进在于没有规定最大的n位数的范围，会不会溢出，也就是说我们要考虑大数问题。最常用也是最容易的方法是用字符串或者数组去表达大数。

bool increment(char* number)
{
bool IsOverFlow = false;
int TakeOver = 0;
int length = strlen(number);

for(int index = length - 1; index >= 0; --index)
{
int num = number[index] - '0' + TakeOver;
if(index == length - 1)
++num;

if(num >= 10)
{
if(index == 0)
IsOverFlow = true;
else
{
num -= 10;
TakeOver = 1;
number[index] = '0' + num;
}
}
else
{
number[index] = '0' + num;
break;
}
}
return IsOverFlow;
}

void PrintNumber(char* number)
{
bool IsBegin0 = true;
int length = strlen(number);
for(int i = 0; i < length; ++i)
{
if(IsBegin0 && number[i] != '0')
IsBegin0 = false;
if(!IsBegin0)
printf("%c", number[i]);
}
}

void Print1ToMaxN(int n)
{
if(n <= 0)
return;
char* number = new char[n+1];
memset(number, '0', n);
number
= '\0';
while(!increment(number))
{
PrintNumber(number);
cout << " " ;
}
cout << endl;
delete[] number;
}

递归解法：

void PrintToMaxOfDigits
{
if(n <= 0)
return;
char* number = new char[n + 1];
number
= '\0';

for(int i = 0; i < 10; ++i)
{
number[0] = i + '0';
PrintToMaxOfDigitsRecursively(number, n, 0);
}
delete[] number;
}

void PrintToMaxOfDigitsRecursively(char* number, int length, int index)
{
if(index == length - 1)
{
PrintNumber(number);
return;
}

for(int i = 0; i < 10; ++i)
{
number[index + 1] = i + '0';
PrintToMaxOfDigitsRecursively(number, length, index + 1);
}
}