Leetcode233 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 Solution233 {
    public int countDigitOne(int n) {
        int ans = 0;
        long cnt = 10;

        if(n <= 0) return ans;

        while (n/cnt >= 0) {
            long next = n/cnt, before = n%(cnt/10), mod = (n-(next*cnt))/(cnt/10);

            ans += next*(cnt/10);

            if(mod == 1) ans += before+1;
            else if(mod > 1) {
                ans += cnt/10;
            }

            cnt *= 10;
            //System.out.println("next = "+next+" before = "+before+" mod = "+mod+" n = "+n+" cnt = "+cnt+" ans ="+ans);

            if(next == 0) break;
        }

        return ans;
    }
    public int compute(int n) {
        int ans = 0;

        while (n > 0) {
            if(n%10 == 1)
                ans += 1;
            n = n / 10;
        }
        return ans;
    }
    public int countOne(int n) {
        int ans = 0;

        for (int i = 1; i <= n; i++) {
            ans+=compute(i);
        }

        return ans;
    }
    public static void main(String[] args) {
        Solution233 result = new Solution233();
        long ans = result.countDigitOne(1410065408);
        //int ans = result.countOne(7925);
        System.out.println(ans);
    }
}