从1到整数n中1出现的次数

题目：输入一个整数n，求从1到n这n个整数的十进制表示中1出现的次数。例如输入12，从1到12这些整数中包含1的数字有1,10,11和12共出现了5次。

 

不考虑时间效率的解法：

int NumberOf1Between1AndN(unsigned int n)
{
 int number=0;
 for(unsigned int i=1;i<=n;++i)
 number +=NumberOf1(i);
 return number;
 }
int NumberOf1(unsigned int n)
{
 int number=0;
 while(n)
 {
  if(n%10==1)
  number++;
  n=n/10;
 }
return number;
}

从数字规律着手明显提高时间效率：

int Numberof1Between1AndN(int n)
{
 if(n<=0)
 return 0;
 char strN[50];
 sprintf(strN,"%d",n);//将整数转换为字符串
 return NumberOf1(strN);
 }

 int NumberOf1(const char* strN)
{
 if(!strN||*strN<'0'||*strN>'9'||*strN=='')
 return 0;
 int first=*strN-'0';
 unsigned int length=static_cast<unsigned int>(strlen(strN));
 if(length==1&&first==0)
 return 0;
 if(length==1&&first>0)
 return 1;

 int numFirstDigit=0;
 if(first>1)
 numberFirstDigit=PowerBase10(length-1);
 else if(first==1)
 numFirstDigit=atoi(strN+1)+1;//将字符串转换为整数
 
 int numOtherDigits=first*(length-1)*PowerBase10(length-2);
 int numRecursive=NumberOf1(strN+1);

 return numFirstDigit+numOtherDigits+numRecursive;
}

int PowerBase10(unsigned int n)
{
 int result=1;
 for(unsigned int i=0;i<n;++i)
 result *=10;
 return result;
 }