leetcode: Number of Digit One

思路是从个位开始一位位的计算每一位1出现的次数....对于每一位来说由三种情况

a=n/m(个位是当前计算的这位)，b=n%m（这位之后的余数）；

那么(a%10)：

case > 1: count+= (a/10+1)*m;

case < 1: count+= (a/10)*m;

case == 1: count+= (a/10)*m+b+1;

public class Solution {
    public int countDigitOne(int n) {
        long m=1,count=0;
        while( m<=n )
        {
            long a=n/m,b=n%m;
            if( a%10<1 )
            {
                count+=(a/10)*m;
            }
            else if( a%10>1 )
            {
                count+=(a/10+1)*m;
            }
            else
            {
                count+=(a/10)*m+b+1;
            }
            m*=10;
        }
        return (int)(count);
    }
}