Understanding when and how to use Java 8 Lambdas

Question

I have been spending some time trying to learn some of Java 8's new features. As an exercise, I wrote a MergeSort using some Java 8 Functional Interfaces. I'm including the full code below (bugs/optimizations may exist, I am only interested in them if they relate to Java 8 features). My question is, I believe there are opportunities to utilize lambda expressions when I am using my functional interfaces, but it's just not clicking in my brain yet. It feels like everytime I am calling apply, there should be a way I can use "->" instead. Can someone please show me the light?

Merge Function written using BinaryOperator Functional Interface

public class Merge implements BinaryOperator<int[]>{

@Override
public int[] apply(int[] t, int[] u) {

    int[] result = new int[t.length + u.length];

    for (int i = 0, j = 0, k = 0; i < result.length; i++){

        if( j == t.length){
            result[i] = u[k++];
        } else if (k == u.length) {
            result[i] = t[j++];
        } else {
            result[i] = t[j] < u [k] ? t[j++] : u[k++];
        }

    }

    return result;
}

}

MergeSort written as a java.util.function.Function

public class MergeSort implements Function<int[], int[]>{

Merge merge = new Merge();

@Override
public int[] apply(int[] t) {

    if(t.length <= 1){
        return t;
    }

     return merge.apply( apply(Arrays.copyOfRange(t, 0, t.length / 2)), 
                         apply(Arrays.copyOfRange(t, t.length / 2, t.length )));

}

}

Main with one simple test case

public class MergeSortMain {

public static void main(String[] args) {

    int values[] = {3,12,6,7,2,1,23,4,5,7,8,4,2,5,365};

    MergeSort mergeSort = new MergeSort();

    System.out.println(Arrays.toString(mergeSort.apply(values)));
}

}

produces

[1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 12, 23, 365]

Answer 1

The idea of lambda expressions is that instead of creating a class that implements a functional interface, you can define a lambda expression of the type of that interface.

For example, your Merge class implements BinaryOperator<int[]> and can be replaced by the following lambda expression :

BinaryOperator<int[]> merge = (t,u) -> {
    int[] result = new int[t.length + u.length];
    for (int i = 0, j = 0, k = 0; i < result.length; i++){
        if( j == t.length){
            result[i] = u[k++];
        } else if (k == u.length) {
            result[i] = t[j++];
        } else {
            result[i] = t[j] < u [k] ? t[j++] : u[k++];
        }
    }
    return result;
};

Now we can similarly create a lambda expression to replace the MergeSort class, and, combining the two lambdas, we get :

public class MergeSortMain {
    public static Function<int[], int[]> mergeSort;
    public static void main(String[] args) {
        int values[] = {3,12,6,7,2,1,23,4,5,7,8,4,2,5,365};
        mergeSort = l -> {
            BinaryOperator<int[]> merge = (t,u) -> {
                int[] result = new int[t.length + u.length];
                for (int i = 0, j = 0, k = 0; i < result.length; i++){
                    if( j == t.length){
                        result[i] = u[k++];
                    } else if (k == u.length) {
                        result[i] = t[j++];
                    } else {
                        result[i] = t[j] < u [k] ? t[j++] : u[k++];
                    }
                }
                return result;
            };
            if(l.length <= 1){
                return l;
            }
            return merge.apply( mergeSort.apply(Arrays.copyOfRange(l, 0, l.length / 2)), 
                                mergeSort.apply(Arrays.copyOfRange(l, l.length / 2, l.length )));
        };
        System.out.println(Arrays.toString(mergeSort.apply(values)));
    }
}

Some points regarding this code :

1. I had to rename the parameter of mergeSort lambda from t to l, since t is also used in the merge lambda.
2. I had to declare the mergeSort lambda as a static member (prior to assigning its value), since it contains recursive calls to itself.