本文共 1564 字，大约阅读时间需要 5 分钟。

例如：输入3，输出 1 ~ 999，必须考虑大数问题

/* 打印从 1到最大的n位数 */bool Increment(char *number){	bool isOverflow = false;	int nTakeOver = 0;	int nLength = strlen(number);	for (int i = nLength-1; i >= 0; i--)	{		int nSum = number[i] - '0' + nTakeOver; //初始置为 0		if (i == nLength - 1) //第一位保证必须是 1			nSum++;		if (nSum >= 10) 		{			if (i == 0)				isOverflow = true;			else			{				nSum -= 10;				nTakeOver = 1;				number[i] = '0' + nSum;			}		}		else		{			number[i] = '0' + nSum;			break;		}	}	return isOverflow;}void PrintNumber(char *number){	bool isBeginning0 = true;	int nLength = strlen(number);	for (int i = 0; i < nLength; ++i)	{		if (isBeginning0 && number[i] != '0') //找到第一位不为0的数字字符			isBeginning0 = false;		if (!isBeginning0)  //从第一位不为0的数字字符开始打印			printf("%c", number[i]);	}	printf("\t");}void PrintToMaxOfNDigits(int n){	if (n <= 0)		return;	char *number = new char[n+1]; //开辟n+1大小的空间	memset(number, '0', n); //前n位置0	number[n] = '\0'; //末尾置为'\0'	while (!Increment(number))		PrintNumber(number);	delete []number;}///* 方法二 */void Print1ToMaxOfNDigitsREcursively(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';		Print1ToMaxOfNDigitsREcursively(number, length, index+1);	}}void Print1ToMaxOfNDigits(int n){	if (n<=0)		return;	char *number = new char[n+1];	number[n] = '\0';	for (int i=0; i<10; ++i)	{		number[0] = i + '0';		Print1ToMaxOfNDigitsREcursively(number, n, 0);	}	delete []number;}void main(){	int n = 3;//	PrintToMaxOfNDigits(n);	Print1ToMaxOfNDigits(n);	}

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