是否有更有效的方法来比较If语句中的3个以上的项目？

Question

I am writing a program for a class assignment and I want to compare multiple items in an array for equivalence. My statement basically looks like:

if (a == b && b == c && c == d && d == e && e == f) {
   // do stuff
}

This condition seems incredibly verbose, and I am wondering if there's a shorter way to write this.

Answer 1

This may appear to be verbose , but at least it is reasonably efficient and clear .

Don't overdo it with optimizing code for brevity.

You could easily make a helper function

boolean allEqual(Object ... objs) {
    for(int i = 1; i < obj.length; i++) {
      if (!objs[i].equals(objs[0])) {
        return false;
      }
    }
    return true;
}

But this will create additional objects; in particular with primitive values:

if (allEqual(1.1,2.1,3.1,4.1))

will create 5 objects, that need to be garbage collected.

Make sure you actually want == and not equals .

Answer 2

My suggestion is to create a set and if its size is one then all elements are equal, otherwise not:

public static < E > boolean areAllEqual( E[] inputArray ){
    Set<E> mySet = new HashSet<E>(Arrays.asList(inputArray));
    return mySet.size() == 1;
}    

Following is complete code example:

import java.util.*;
public class HelloWorld{
    public static < E > boolean areAllEqual( E[] inputArray ){
        Set<E> mySet = new HashSet<E>(Arrays.asList(inputArray));
        return mySet.size() == 1;
    }    
     public static void main(String []args){
        Integer[] intArray = { 1, 2, 3, 4, 5 };
        Double[] doubleArray = { 1.1, 2.2, 3.3, 4.4 };
        Character[] charArray = { 'H', 'H', 'H', 'H', 'H' };

        System.out.println("It should work");
        if (areAllEqual(intArray))
            System.out.println("inarray Yes");
        else
            System.out.println("inarray No");

        if (areAllEqual(doubleArray))
            System.out.println("doubleArray Yes");
        else
            System.out.println("doubleArray No");

        if(areAllEqual(charArray))
            System.out.println("charArray  Yes");
        else
            System.out.println("charArray  No");                            
     }
}

output:

It should work
inarray No
doubleArray No
charArray  Yes  // all elements in char array are eq

Answer 3

This could be done in Java 8 the following, given that you already have an array:

int[] numbers = {1, 2, 3, 4, 5};
if (Arrays.stream(numbers).allMatch(i -> (i == numbers[0]))) {
    //they are equal
}

What it does is:

• It encapsulates your array in an IntStream (it does not create extra objects, apart from the IntStream )
• It checks if all objects return true on the IntPredicate .
• The IntPredicate is i -> (i == numbers[0]) , which maps i to (i == numbers[0]) .

It would be advised to encapsulate this in a helper method, and you'd want to early bail if the array is empty, like this:

public static boolean allEquals(final int[] input) {
    return (input.length > 0 && Arrays.stream(input).allMatch(i -> i == input[0]));
}

This depends though on whether it should return true or false if there's no input... Is everything equal then?

Please note that you'd need to make this code for all primitives you are using. As an extra, you can also provide a generic version:

public static <T> boolean allEquals(final T[] input) {
    return (input.length > 0 && Arrays.stream(input).allMatch(i -> i.equals(input[0])));
}

Now when you do not have a suitable structure, meaning they are most likely just variables, then you will need to decide whether you are needing the extra performance or can live with the overhead of passing them in as an varargs input, that implicitely creates an array, the code for that would be really similar:

public static <T> boolean allEquals(final T... input) {
    return (input.length > 0 && Arrays.stream(input).allMatch(i -> i.equals(input[0])));
}

You can almost certainly live with the extra overhead, only in very performance critical applications, you need to wary if your method gets executed a load of times.