如何从应用程序中删除大量的条件语句？

huangapple
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英文:

How to remove the amounts of ifs from the application?

问题

以下是简化后的代码：

  1. private static final Scanner scanner = new Scanner(System.in);
  3. public static void main(String[] args) {
  4.     int N = scanner.nextInt();
  5.     scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
  6.     scanner.close();
  7.     
  8.     if (N % 2 == 1 || (N >= 6 && N <= 20)) {
  9.         System.out.println("Weird");
  10.     } else if (N >= 2 && N <= 5) {
  11.         System.out.println("Not Weird");
  12.     } else if (N > 20) {
  13.         System.out.println("Not Weird");
  14.     }
  15. }

这个简化后的代码只使用了两个条件语句，分别处理"Weird"和"Not Weird"的情况，消除了多余的嵌套条件。

英文:

The code I say has to complete these assignments:
Given an integer,N, perform the following conditional actions:

If N is odd, print Weird
If N is even and in the inclusive range of 2 to 5 , print Not Weird
If N is even and in the inclusive range of 6 to 20 , print Weird
If N is even and greater than 20 , print Not Weird
Complete the stub code provided in your editor to print whether or not N is weird.

My code looked like this:

  1. private static final Scanner scanner = new Scanner(System.in);
  3.     public static void main(String[] args) {
  4.         int N = scanner.nextInt();
  5.         scanner.skip(&quot;(\r\n|[\n\r\u2028\u2029\u0085])?&quot;);
  6.         scanner.close();
  7.         int Numberparorimpa = N % 2;
  8.         if(N &lt; 2 || Numberparorimpa ==1 || N &lt;=20 &amp;&amp;  N &gt;=6 ){
  9.             System.out.println(&quot;Weird&quot;);
  10.         }else{
  11.             if(N &gt;=2 &amp;&amp; Numberparorimpa == 0){
  12.                  System.out.println(&quot;Not Weird&quot;);
  13.             }else{
  14.              if(Numberparorimpa == 0 &amp;&amp; N &gt;=6 || N&lt;=20){
  15.                   System.out.println(&quot;Weird&quot;);
  16.              }else{
  17.                  if(Numberparorimpa== 0 &amp;&amp; N&gt; 20){
  18.                      System.out.println(&quot;Not Weird&quot;);
  19.                  }else{
  20.                      return;
  21.                  }
  22.              }   
  23.             }
  24.         }
  25.             }
  26.         
  27.     }

How can I reduce the IFs of this code?

答案1

得分: 1

如果N是奇数或在6到20的范围内，它就是奇怪的。

否则，N要么是偶数，要么不在范围内，所以它不奇怪。

英文:

I think an optimize version could be this :

  1. if (N % 2 == 1 || (N &gt;= 6 &amp;&amp; N &lt;= 20)) {
  2.   System.out.println(&quot;Weird&quot;);
  3. }
  4. else {
  5.   System.out.println(&quot;Not Weird&quot;);
  6. }

If N is odd or N in range of 6 to 20 it's weird.

Else N is either even or not in the range so it's not weird.

答案2

得分: 1

你可以通过将逻辑提取到方法/类等方式来减少代码的复杂性并提高可读性。此外，嵌套条件很难阅读，应该避免使用。

示例：

  1. public static void main(String[] args) {
  3.     int n = scanner.nextInt();
  4.     scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
  5.     scanner.close();
  7.     String print = isGivenNumberWeird(n) ? "Weird" : "Not Weird";
  8.     System.out.println(print);
  9. }
  11. private static boolean isGivenNumberWeird(int n) {
  12.     boolean isOdd = n % 2 == 1;
  13.     if (isOdd) {
  14.         return true;
  15.     }
  16.     if (n >= 2 && n <= 5) {
  17.         return false;
  18.     }
  19.     if (n >= 6 && n <= 20) {
  20.         return true;
  21.     }
  22.     if (n > 20) {
  23.         return false;
  24.     }
  25. }
英文:

You can reduce the complexity and improve the readability of your code by extracting your logic to method/class etc. Also, nested conditions are hard to read, you should avoid it.

Example:

  1. public static void main(String[] args) {
  3. 	int n = scanner.nextInt();
  4.     scanner.skip(&quot;(\r\n|[\n\r\u2028\u2029\u0085])?&quot;);
  5.     scanner.close();
  7.     String print = isGivenNumberWeird(n) ? &quot;Weird&quot; : &quot;Not Weird&quot;
  8. 	System.out.println(print);
  9. }
  11. private static boolean isGivenNumberWeird(int n) {
  12. 	boolean isOdd = n % 2 == 1;
  13.     if (isOdd) {
  14. 		return true;
  15.     }
  16.     if (n &gt;= 2 &amp;&amp; n &lt;=5) {
  17.     	return false;
  18.     }
  19.     if (n &gt;= 6 &amp;&amp; n &lt;=20) {
  20.     	return true;
  21.     }
  22.     if (n &gt; 20) {
  23.     	return false;
  24.     }
  25. }

huangapple
  • 本文由 发表于 2023年1月9日 16:32:50
  • 转载请务必保留本文链接：https://go.coder-hub.com/75054751.html
  • code-cleanup
  • java
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