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lintcode-medium-Digit Counts

时间:2016-03-18 13:24:00      阅读:262      评论:0      收藏:0      [点我收藏+]

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

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

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) {
        // write your code here
        
        int count = 0;
        
        for(int i = k ; i <= n; i++){
            count += singleCount(i, k);
        }
        
        return count;
    }
    
    public int singleCount(int num, int k){
        if(num == 0 && k == 0)
            return 1;
        
        int count = 0;
        
        while(num > 0){
            if(num % 10 == k)
                count++;
            
            num /= 10;
        }
        
        return count;
    }
    
};

 

lintcode-medium-Digit Counts

原文:http://www.cnblogs.com/goblinengineer/p/5291643.html

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