打印1到最大的n位数

1.递归的方式

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void printNumber(char *number)
{
	int i;

	for (i=0; number[i]!='\0' ; ++i)
	{
		if (number[i] != '0')
		{
			break;
		}
	}

	if (number[i] != '\0')
	{
		printf("%s\t", number+i); /*这个比上面的一个个的打印快了4倍以上*/
	}
}

void print1ToMaxofNDigitsRecurisively(char *number, int n, int index)
{
	int i;

	/*递归结束的条件，已经设置了数字的最后一位*/
	if (index == n)
	{
		printNumber(number);
		return ;
	}

	for (i=0; i<10; ++i)
	{
		number[index] = i + '0';
		print1ToMaxofNDigitsRecurisively(number, n, index+1);
	}
}

void print1ToMaxofNDigits(int n)
{
	char *number;

	if (n <= 0)
	{
		return ;
	}

	number = (char *)malloc(sizeof(char)*(n+1));
	number[n] = '\0';

	print1ToMaxofNDigitsRecurisively(number, n, 0);
	printf("\n");

	free(number);
	number = NULL;
}

int main(void)
{
	int n;

	while (scanf("%d", &n) != EOF)
	{
		print1ToMaxofNDigits(n);
	}
	
	return 0;
}

 

2.字符串模拟数字加法

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void printNumber(char *number)
{
	int i;

	for (i=0; number[i]!='\0' ; ++i)
	{
		if (number[i] != '0')
		{
			break;
		}
	}

/*	for (; number[i]!='\0'; ++i)
	{
		printf("%c", number[i]);
	}
*/
	printf("%s\t", number+i); /*这个比上面的一个个的打印快了4倍以上*/
}

int increment(char *number, int n)
{
	int i, num;
	int isOverflow = 0;
	int carry = 0;

	for (i=n-1; i>=0; --i)
	{
		num = number[i] - '0' + carry;

		if (i == n-1)
		{
			++num; /*只有个位数执行加1操作*/
		}

		if (num > 9)
		{
			if (i == 0)
			{
				isOverflow = 1;
				break;
			}
			else
			{
				num -= 10;
				carry = 1;
				number[i] = num + '0';
			}
		}
		else
		{
			number[i] = num + '0';
			break; /*没有进位时结束*/
		}
	}

	return isOverflow;
}

void print1ToMaxofNDigits(int n)
{
	char *number;

	if (n <= 0)
	{
		return ;
	}

	number = (char *)malloc(sizeof(char)*(n+1));
	memset(number, '0', sizeof(char)*(n));
	number[n] = '\0';

	while (0 == increment(number, n))
	{
		printNumber(number);
	}
	printf("\n");

	free(number);
	number = NULL;
}

int main(void)
{
	int n;

	while (scanf("%d", &n) != EOF)
	{
		print1ToMaxofNDigits(n);
	}
	
	return 0;
}