1的数目

　　给定一个十进制数N，写下从1开始，到N的所有整数，然后数一下其中出现的所有”1“的个数。

例如N=2，写下1，2。这样只出现了1个”1“.

public int getNumberOfOne(int n){
        int factor =1 ;
        int count = 0;
        while(n/factor!=0){
            int lowerNumber = n-(n/factor)*factor;
            int currentNumber = (n/factor)%10;
            int highNumber = n/(factor*10);
            switch(currentNumber){
            case 0:
                count += highNumber*factor;
                break;
            case 1:
                count+=highNumber*factor+lowerNumber+1;
                break;
            default:
                count+=(highNumber+1)*factor;
                break;
            }
            factor *= 10;
        }
        return count;
    }