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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;    }

 

1的数目