[LeetCode] 2269. Find the K-Beauty of a Number

The k-beauty of an integer num is defined as the number of substrings of num when it is read as a string that meet the following conditions:

It has a length of k.
It is a divisor of num.
Given integers num and k, return the k-beauty of num.

Note:
Leading zeros are allowed.
0 is not a divisor of any value.
A substring is a contiguous sequence of characters in a string.

Example 1:
Input: num = 240, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:

• “24” from “240”: 24 is a divisor of 240.
• “40” from “240”: 40 is a divisor of 240.
  Therefore, the k-beauty is 2.

Example 2:
Input: num = 430043, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:

• “43” from “430043”: 43 is a divisor of 430043.
• “30” from “430043”: 30 is not a divisor of 430043.
• “00” from “430043”: 0 is not a divisor of 430043.
• “04” from “430043”: 4 is not a divisor of 430043.
• “43” from “430043”: 43 is a divisor of 430043.
  Therefore, the k-beauty is 2.

Constraints:
1 <= num <= 109
1 <= k <= num.length (taking num as a string)

找到一个数字的 K 美丽值。

一个整数 num 的 k 美丽值定义为 num 中符合以下条件的 子字符串 数目：

子字符串长度为 k 。
子字符串能整除 num 。
给你整数 num 和 k ，请你返回 num 的 k 美丽值。

注意：
允许有 前缀 0 。
0 不能整除任何值。
一个 子字符串 是一个字符串里的连续一段字符序列。

思路

将 num 转换为字符串，方便提取长度为 k 的子串。
遍历所有可能的 k 长度子串，将其转换为整数。
判断该整数是否能整除 num（需要排除除数为 0 的情况）。
计数满足条件的子串数量。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
	public int divisorSubstrings(int num, int k) {
		int count = 0;
		String numStr = String.valueOf(num);
		for (int i = 0; i <= numStr.length() - k; i++) {
			int curNum = Integer.parseInt(numStr.substring(i, i + k));
			// System.out.println(curNum);
			if (curNum != 0 && num % curNum == 0) {
				count++;
			}
		}
		return count;
	}
}