1到n中1的个数

读《剑指Offer》有感，记下下面内容以备忘记：
#include <iostream>
using namespace std;

int Numof1ToN(int n)
{
	int res = 0;
	for (int i = 1;i <= n; ++i)
	{
		int j = i;
		while(j)
		{
			if(j % 10 == 1)
			{
				res++;
			}
			j /= 10;
		}
	}

	return res;
}

int Power(int base, int n)
{

	if (n == 0) return 1;

	else if (n == 1) return base;

	int half = Power(base, n / 2);

	return n % 2 == 0? half * half:half * half * base;

}
//思路：21345
// 1- 5, 6-45, 46 - 345, 346-1345, 1346-21345
//拆成5部分：
//每一部分：1）若最高位为1，则firstNum = 除了最高位外的数 + 1 比如：1345 则firstnum = 345 + 1;
//          2）若最高位不为1， 则firstNum = 10^(len - 1)
//除了最高位，其他位上OtherNum = first * (len - 1) * 10^(len - 2);

int Numof1ToNLogN(char * str)
{
	if(!str || *str < '0' || *str > '9' || *str == 0)
	{
		return 0;
	}

	int first= *str - '0';
	int len = strlen(str);

	if (len == 1 && first ==0)
	{
		return 0;
	}

	if (len == 1&& first > 0)
	{
		return 1;
	}

	int numFirst = 0;
	if(first > 1)
	{
		numFirst = Power(10, len - 1);
	}
	else if (first == 1)
	{
		numFirst = atoi(str + 1) + 1;
	}

	int numOther = first * (len - 1) * Power(10, len - 2);
	int numRecursin = Numof1ToNLogN(str + 1);
	return numFirst + numOther + numRecursin;

}

int main()
{
	printf("NUM: %d\n", Numof1ToN(121));
	printf("NUMlogn: %d\n", Numof1ToNLogN("121"));
	return 0;
}