Question

In this assignment, you sort a list of strings using mergesort and the compareToIgnoreCase method of the String class.

The strings are provided as arguments to your main method. In case you haven't yet learned to configure command-line arguments to main in your IDE, now is a good time to learn.

Your program should sort the array of strings using mergesort, then print the strings one per line, in order from smallest ("a") to largest ("z"). The name of your class should be MergeSort.

Example

% java MergeSort z x y w
w
x
y
z

Choices

Your mergesort must have a recursive implementation. That is, you must have a method merge that, in its recursive case, calls itself twice.

On the other hand, you can choose the details of how to implement mergesort. In particular, you may use different indices to indicate subarrays, and have an input and output array. Or, you may copy portions of the array to be sorted, then merge the results. In other words, you can use static storage -- one input array and one output (or spare) array -- or you can use dynamic storage -- subarrays are allocated as needed.

Your merge method may return void, or may return a sorted array -- again, this is your choice.

One thing you are NOT allowed to do is to call a built-in sort method.

Answer 1

main.java

import java.util.*;

public class Main {
public static void main(String[] args) {
String[] arr = { "Akshay", "Anthony", "Angelina",
"George", "Victor", "Ankit", "Chama" };
System.out.println("Array after Sorting: ");
String[] arr1 = sorting(arr);
for (int u = 0; u < arr1.length; u++) {
System.out.println(arr1[u] + " ");
}

}

public static String[] sorting(String[] arr2) {
String [] arr3 = new String[arr2.length];
if (arr2.length == 1) {
arr3 = arr2;
} else {
int middle = arr2.length/2;
String[] arr4 = null;
String[] arr5 = null;
if ((arr2.length % 2) == 0) {
arr4 = new String[arr2.length/2];
arr5 = new String[arr2.length/2];
} else {
arr4 = new String[arr2.length/2];
arr5 = new String[(arr2.length/2)+1];
}
int u=0;
int v=0;
for ( ; u < middle; u++) {
arr4[u] = arr2[u];
}
for ( ; u < arr2.length; u++) {
arr5[v++] = arr2[u];
}
arr4 = sorting(arr4);
arr5 = sorting(arr5);
arr3 = merging(arr4,arr5);
}

return arr3;
}

private static String[] merging(String[] arr4, String[] arr5) {
String[] arr6 = new String[arr4.length+arr5.length];
int n = 0;
int n1 = 0;
int n2 = 0;
int n3 = 0;
while (n < arr4.length || n1 < arr5.length) {
if (n == arr4.length) {
arr6[n2++] = arr5[n1++];
} else if (n1 == arr5.length) {
arr6[n2++] = arr4[n++];
} else {  
n3 = arr4[n].compareTo(arr5[n1]);
if (n3 > 0) {
arr6[n2++] = arr5[n1++];
} else if (n3 < 0) {
arr6[n2++] = arr4[n++];
} else {
arr6[n2++] = arr4[n++];
}
}
}
return arr6;
}

}

Output:

java version "1.8.0 31" ava(TM) SE Runtime Environment (build 1.8.031-b13) Java HotSpot (TM) 64-Bit Server VM (build 25.31-be7, mixed mode) import java.util.*; - public class Main Array after Sorting Akshay Angelina Ankit Anthony Chama George Victor public static void main(String[] args) [ String[] arr "Akshay" "Anthony", "Angelina" "George", "Victor", "Ankit", "Chama"); system.out.print1n("Array after sorting: "의 Stringl] arri - sorting(arr); for (intu-e; u <arri.1ength; u) System.out.println(arri[u] ) public static String[] sorting (String[] arr2) String [] arr3 new String[arr2.1ength]; İf (arr2.length-- 1) { arr3 arr2; else int middle arr2.length/2; String[] arr4 null; String[] arr5 = null; if ((arr2. length % 2) 0) arr4 new String[arr2.length/2]; arrS = new String[arr2. length/2); arr4 = new String[arr2. length/2); arrS = new String[ (arr2. length/2)+1]; int u-e int v 8; foru< middle; u+) arr4(u) = arr2(u);

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Hope this helps....