【codewars】Are they the "same"

原贴地址
https://www.codewars.com/kata/550498447451fbbd7600041c/train/java

简介如下：
Given two arrays a and b write a function comp(a, b) (compSame(a, b) in Clojure) that checks whether the two arrays have the "same" elements, with the same multiplicities. "Same" means, here, that the elements in b are the elements in a squared, regardless of the order.

Examples

Valid arrays

a = [121, 144, 19, 161, 19, 144, 19, 11]
b = [121, 14641, 20736, 361, 25921, 361, 20736, 361]
comp(a, b) returns true because in b 121 is the square of 11, 14641 is the square of 121, 20736 the square of 144, 361 the square of 19, 25921 the square of 161, and so on. It gets obvious if we write b's elements in terms of squares:

a = [121, 144, 19, 161, 19, 144, 19, 11]
b = [1111, 121121, 144144, 1919, 161161, 1919, 144144, 1919]
Invalid arrays

If we change the first number to something else, comp may not return true anymore:

a = [121, 144, 19, 161, 19, 144, 19, 11]
b = [132, 14641, 20736, 361, 25921, 361, 20736, 361]
comp(a,b) returns false because in b 132 is not the square of any number of a.

a = [121, 144, 19, 161, 19, 144, 19, 11]
b = [121, 14641, 20736, 36100, 25921, 361, 20736, 361]
comp(a,b) returns false because in b 36100 is not the square of any number of a.

Remarks

a or b might be [] (all languages except R, Shell). a or b might be nil or null or None (except in Haskell, Elixir, C++, Rust, R, Shell).

If a or b are nil (or null or None), the problem doesn't make sense so return false.

If a or b are empty the result is evident by itself.

Note for C

The two arrays have the same size (> 0) given as parameter in function comp.

方法类

public class AreSame {
    
    public static boolean comp(int[] a, int[] b) {
    return null;
  }
}

测试类

import static org.junit.Assert.*;
import org.junit.Test;


public class AreSameTest {

    @Test
    public void test1() {
        int[] a = new int[]{121, 144, 19, 161, 19, 144, 19, 11};
        int[] b = new int[]{121, 14641, 20736, 361, 25921, 361, 20736, 361};
        assertEquals(AreSame.comp(a, b), true); 
    }
}

我的答案

import java.util.ArrayList;
import java.util.List;
import java.util.Arrays;
public class AreSame {
    
    public static boolean comp(int[] a, int[] b) {
        System.out.println(Arrays.toString(a));
        System.out.println(Arrays.toString(b));
        
        if (a == null || b == null || a.length != b.length ) {
            return false;
        }
        
        
        List blist = new ArrayList<>();
        for (int i = 0; i < b.length; i++) {
            blist.add(b[i]);
        }
        
        boolean flag = true;
        for(int i = 0; i < a.length; i++){
            int tempa = a[i];
            int square = tempa * tempa;
            if (!blist.contains(square)) {
                flag = false;
                break;
            }else {
                blist.remove(Integer.valueOf(square));
            }
        }
        
        return flag;
  }
}

solution 地址
https://www.codewars.com/kata/550498447451fbbd7600041c/solutions/java

best prictice

import java.util.Arrays;

public class AreSame {
  public static boolean comp(final int[] a, final int[] b) {
    return a != null && b != null && a.length == b.length && Arrays.equals(Arrays.stream(a).map(i -> i * i).sorted().toArray(), Arrays.stream(b).sorted().toArray());
  }
}