LintCode - Digit Counts

DigitCounts.java

// solution 1
public int digitCounts(int k, int n) {
    // write your code here
    if(k < 0 || n < 0) return 0;
    if(k == 0 && n == 0) return 1;
    int factor = 1, count = 0;
    
    while(factor <= n){
        
        int low = n % factor;
        int high = n / (10 * factor);
        int digit = n / factor % 10;
        
        if(digit == k){
            count += high * factor + (k == 0 && high == 0 ? 0 : (low + 1));
        } else if(digit < k) {
            count += high * factor;
        } else if(digit > k){
            count += high * factor + (k == 0 && high == 0 ? 0 : factor);
        }
        
        factor *= 10;
    }
    
    return count;
}

// solution 2
public int digitCounts(int k, int n) {
    // write your code here
    int cnt = 0;
    for (int i = k; i <= n; i++) {
        cnt += singleCount(i, k);
    }
    return cnt;
}

public int singleCount(int i, int k) {
    if (i == 0 && k == 0)
        return 1;
    int cnt = 0;
    while (i > 0) {
        if (i % 10 == k) {
            cnt++;
        }
        i = i / 10;
    }
    return cnt;
}