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.

 

public class Solution {
    static int[] sta;
    public int countDigitOne(int n) {
        if(n<=0) return 0;
        int result = calOnes(n);
        return result;
        
    }
    
    //计算位数
    public int totalPos(int n){
        int count = 0;
        while(n>0){
            count++;
            n /= 10;
        }
        return count;
    }
    //计算10的n次方
    public int times(int n){
        int result = 1;
        while(n>0){
            result *= 10;
            n--;
        }
        return result;
    }
    //计算n位数最大值以内的所有数有多少个1,存在sta[n]中
    public void cal(int len){
        for(int i=1; i<=len; i++){
            sta[i] = times(i-1) + 10*sta[i-1];
        }
        
    }
    //计算1的个数
    public int calOnes(int n){
        int count = totalPos(n);
        sta = new int[count+1];
        
        cal(count);
        
        int result = 0;
        int x=count;
        int i = 0;
        int hPos = 0;
        int rest = n;
        while(x>0){
            hPos = rest/times(x-1);
            rest = rest%times(x-1);
            for(i=0; i<hPos; i++){
                if(i==1){
                    result += (times(x-1) + sta[x-1]);
                }else{
                    result += sta[x-1];
                }
            }
            if(hPos!=0 && i==1) result += (rest+1);
            if(rest>0) x--;
            else x=0;
        }
        
        return result;
    }
}