(*medium)LeetCode 233.Number of Digit One

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

代码如下：

public class Solution {
    public int countDigitOne(int n) {
        if(n<=0)
          return 0;
        int j=1,k=n,constant=0;
        while(k>=10){
            k/=10;
            j*=10;
        }
        if(k>1){
            constant=j;
        }else{
            constant=n%j+1;
        }
        return constant+k*countDigitOne(j-1)+countDigitOne(n%j);
    }
}

　运行结果：