Digit Counts

Posted 北叶青藤

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Digit Counts相关的知识,希望对你有一定的参考价值。

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

Example

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)

分析:

利用while 循环 + num % 10可以获取 num中的所有数字。

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 totalCount = 0;
        for (int i = 0; i <= n; i++) {
            totalCount += counts(k, i);
        }
        return totalCount;
    }
    
    public int counts(int k, int value) {
        int count = 0;
        if (value == 0 && k == 0) return 1;
        while (value != 0) {
            int remainder = value % 10;
            if (remainder == k) {
                count++;
            }
            value = value / 10;
        }
        return count;
    }
};

 

以上是关于Digit Counts的主要内容,如果未能解决你的问题,请参考以下文章

[Algorithm] 3. Digit Counts

Lintcode003.Digit Counts

lintcode-medium-Digit Counts

Project Euler 63: Powerful digit counts

2021-11-04:计算右侧小于当前元素的个数。给你`一个整数数组 nums ,按要求返回一个新数组 counts 。数组 counts 有该性质: counts[i] 的值是 nums[i] 右(

迭代列列表以打印出.value_counts()