public class Solution {
    public int countDigitOne(int n) {
        if(n<0){
            return 0;
        }
        int res =0;
        for(int i=1;i<=n;i++){
            int[] nums=numToArray(i);
            res +=countOne(nums);
        }
        return res;
    }
    
    private int[] numToArray(int n){
        int i=0;
        int tmp = n;
        while(n>0){
            i++;
            n = n/10;
        }
        int[] res= new int[i];
        int index = i-1;
        while(tmp>0){
            res[index--] = tmp%10;
            tmp = tmp/10;
        }
        return res;
    }
    private int countOne(int[] n){
        int res = 0;
        for(int i=0;i<n.length;i++){
            if(n[i] == 1){
                res+=1;
            }
        }
        return res;
    }
    public int countDigitOne2(int n) {
    
        if (n <= 0) return 0;
        int q = n, x = 1, ans = 0;
        do {
            int digit = q % 10;
            q /= 10;
            ans += q * x;
            if (digit == 1) ans += n % x + 1;
            if (digit >  1) ans += x;
            x *= 10;
        } while (q > 0);
        return ans;
    
    }

}
@@ -1145,8 +1145,330 @@
     %7D%0A%0A%7D
+%0A%0A/*The idea is to calculate occurrence of 1 on every digit. There are 3 scenarios, for example%0A%0Aif n = xyzdabc%0Aand we are considering the occurrence of one on thousand, it should be:%0A%0A(1) xyz * 1000                     if d == 0%0A(2) xyz * 1000 + abc + 1           if d == 1%0A(3) xyz * 1000 + 1000              if d %3E 1%0A%0A*/

Add describtion.