LintCode-Digit Counts

Count the number of k's between 0 and n. k can be 0 - 9.

Example

if n=12, in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], we have FIVE 1's (1, 10, 11, 12)

Analysis:

Method 1: directly count every number.

Method 2: Analytical solution.

Every time, calculate how many k has appears on a specific postion, i.e., on 1, 10, 100,.....

Solution 1:

class Solution {
    /*
     * param k : As description.
     * param n : As description.
     * return: An integer denote the count of digit k in 1..n
     */
    public int digitCounts(int k, int n) {
        int[] record = new int[10];
        Arrays.fill(record,0);
        for (int i=0;i<=n;i++){
            String temp = Integer.toString(i);
            for (int j=0;j<temp.length();j++){
                int ind = (int) (temp.charAt(j)-'0');
                record[ind]++;
            }
        }
        return record[k];
                
    }
};

Solution 2:

class Solution {
    /*
     * param k : As description.
     * param n : As description.
     * return: An integer denote the count of digit k in 1..n
     */
    public int digitCounts(int k, int n) {
        int res = 0;
        int base = 1;
        while (base<=n){
            int part1 = n/(base*10);
            if (base>1 && k==0 && part1>0) part1--;
            part1 *= base;
            int bar = n/base%10;
            int part2 = 0;
            if (k<bar) part2 = base;
            else if (k==bar) part2 = n%base+1;
            if (k==0 && n<base*10) part2 = 0;
            res += part1+part2;
            base*=10;
        }
        return res;    
    }
};