英文:
How to remove the amounts of ifs from the application?
问题
以下是简化后的代码：
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int N = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
scanner.close();
if (N % 2 == 1 || (N >= 6 && N <= 20)) {
System.out.println("Weird");
} else if (N >= 2 && N <= 5) {
System.out.println("Not Weird");
} else if (N > 20) {
System.out.println("Not Weird");
}
}
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:
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int N = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
scanner.close();
int Numberparorimpa = N % 2;
if(N < 2 || Numberparorimpa ==1 || N <=20 && N >=6 ){
System.out.println("Weird");
}else{
if(N >=2 && Numberparorimpa == 0){
System.out.println("Not Weird");
}else{
if(Numberparorimpa == 0 && N >=6 || N<=20){
System.out.println("Weird");
}else{
if(Numberparorimpa== 0 && N> 20){
System.out.println("Not Weird");
}else{
return;
}
}
}
}
}
}
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 :
if (N % 2 == 1 || (N >= 6 && N <= 20)) {
System.out.println("Weird");
}
else {
System.out.println("Not Weird");
}
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：
public static void main(String[] args) {
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
scanner.close();
String print = isGivenNumberWeird(n) ? "Weird" : "Not Weird";
System.out.println(print);
}
private static boolean isGivenNumberWeird(int n) {
boolean isOdd = n % 2 == 1;
if (isOdd) {
return true;
}
if (n >= 2 && n <= 5) {
return false;
}
if (n >= 6 && n <= 20) {
return true;
}
if (n > 20) {
return false;
}
}
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:
public static void main(String[] args) {
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
scanner.close();
String print = isGivenNumberWeird(n) ? "Weird" : "Not Weird"
System.out.println(print);
}
private static boolean isGivenNumberWeird(int n) {
boolean isOdd = n % 2 == 1;
if (isOdd) {
return true;
}
if (n >= 2 && n <=5) {
return false;
}
if (n >= 6 && n <=20) {
return true;
}
if (n > 20) {
return false;
}
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论