剑指offer-43

输入整数N，求1~n这N个十进制数中1出现的次数

解法一：粗暴遍历，再计算每个数字包含1的个数，总数累加，O（n），有些数字没有必要走，比如22~30不可能出现1

解法二：N变成数组，观察各个位数的特点，递归地拆分最高位和低位两部分，计算1出现的情况，反应不够快的话，就背吧

解法一：
int getNumOf1(int n)//diy
{
	int numof1 = 0;
	while (n>0)
	{
		if (n % 10 == 1)
			numof1++;
		n /= 10;
	}
	return numof1;
}

int NumberOf1Between1AndN_Solution3(int n)//diy
{
	if (n <= 0) return 0;
	int NumOf1 = 0;
	for (int i = 1; i <= n; i++)
		NumOf1 += getNumOf1(i);
	return NumOf1;
}

int getPowerBase10(int exponent)
{
	if (exponent <= 0) return 1;
	int ret = 10;
	for (int i = 1; i < exponent; i++)
		ret *= 10;
	return ret;
}

解法二：
int getNumOf1(int* Arr, int length, int currPos)//diy
{
	if (currPos >= length) return 0;
	if (currPos==length-1)
	{
		if (Arr[currPos]<1)
			return 0;
		else
			return 1;
	}

	int sumof1 = 0;
	if (Arr[currPos]>1)
		sumof1 += getPowerBase10(length - currPos-1);
	else if (Arr[currPos]==1)
	{
		for (int i = length - 1; i > currPos;i--)
			sumof1 += Arr[i] * getPowerBase10(length - i - 1);
		sumof1++;
	}
	
	sumof1 += Arr[currPos] * (length - currPos - 1)*getPowerBase10(length - currPos - 2);
	sumof1 += getNumOf1(Arr, length, currPos + 1);
	return sumof1;
}

int NumberOf1Between1AndN_Solution4(int n)//diy
{
	if (n <= 0) return 0;
	int len = 0;
	for (int m = n; m > 0; m /= 10)
		len++;
	int* Arr = new int[len]();
	int temp = n;
	for (int i = len - 1; i >= 0;i--)
	{
		Arr[i] = temp % 10;
		temp /= 10;
	}
	return getNumOf1(Arr, len, 0);
}