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

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敘述

之前寫了兩題比較難的 Sliding Window ，覺得自己還沒有到很熟悉，所以找了幾題來練習。

  • 難度: Easy
  • 花費時間: 30 min
  • 題目

給你兩個數字 num 跟 k ，前者是你要處理的數字，後者是 substring 的寬度，
依照 Sliding Window 的原理，在 num 裡取出所有大小為 k 的 substring ，回傳這些值之中可以整除 num 的值的數量。

點我開啟限制與範例 
限制:


    

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



Example 1:


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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:


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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.


Example 3:


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Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

筆記

這題很明顯的告訴你他用到 Sliding Window
並且思路也不難，基本上算是練手題。

  1. 設定 window 左邊，右邊基本上就是左邊 + k ，所以也不用設。
  2. 把每個當下的 window 拿出來跟 num 比，沒有餘數就可以把 count + 1 了

程式碼

java

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public int divisorSubstrings(int num, int k) {
  String numString = String.valueOf(num);
  int left = 0; 
  int window = 0;
  int count = 0;

  while (left + k < numString.length() + 1) {
    for (int i = left; i < left + k; i++) {
      window = window * 10 + Integer.parseInt(String.valueOf(numString.charAt(i)));
    }
    if (window != 0 && num % window == 0) count++;
    left++;
    window = 0;
  }
  return (count);
}

TS

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function divisorSubstrings(num: number, k: number): number {
    let left = 0;
    let window: number = 0;
    let count = 0;

    while ((left + k) < num.toString().length + 1) {
        for (let i = left; i < (left + k); i++) {
            window = window * 10 + parseInt(num.toString().charAt(i));
        }
        if (num % window === 0) count++;
        console.log(window);
        left++;
        window = 0;
    }
    return count;
};

成績

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