merge sort is duplicatiing array entries - java

I was trying to implement in Java the merge sort algorithm according to Cormen's Introduction to Algorithms. The problem with my code (below) is that the main array is duplicating some of its entries during the merge step.
Is someone able to catch what I'm doing wrong?
Thank you!
static void merge(int a[], int p, int q, int r)
{
int n1 = q - p;
int n2 = (r - q);
int [] left = new int[n1 + 1];
int [] right = new int[n2 + 1];
int pp = p;
int qq = q;
for(int i = 0; i < n1; i++)
{
left[i] = a[++pp];
}
for(int i = 0; i < n2; i++)
{
right[i] = a[++qq];
}
left[left.length-1] = Integer.MAX_VALUE;
right[right.length-1] = Integer.MAX_VALUE;
int i = 0;
int j = 0;
for(int k = p; k < r; k++)
{
if(left[i] <= right[j])
{
a[k] = left[i];
i++;
}
else
{
a[k] = right[j];
j++;
}
}
}
static int [] mergeSort(int a[], int p, int r)
{
if(p < r)
{
int q = (p + r)/2;
mergeSort(a, 1, q);
mergeSort(a, q + 1, r);
merge(a, p, q, r);
}
return a;
}

Part of the issue here is the example from the book apparently uses index range from 1 to length. It will be simpler if you change the index range from 0 to length-1, which I assume in the rest of my answer.
Use post increment while copying to left[] and right[] as answered by laune (since index range 0 to length-1).
left[i] = a[pp++];
...
right[i] = a[qq++];
The main issue is the merge function is not checking to see if it reached the end of the left or right run during a merge. This can be fixed by changing the inner if to:
if (i < n1 && (j >= n2 || left[i] <= right[j]))
The recursive calls to merge sort should be:
mergeSort(a, p, q);
mergeSort(a, q, r);
Not shown, but the initial call to mergeSort should be:
mergeSort(a, 0, a.length);
There's no need to allocate the extra element in left and right (since index range is 0 to length-1).
int [] left = new int[n1];
int [] right = new int[n2];

I think that this is in error (as well as its sibling in the next loop):
left[i] = a[++pp];
You want to copy starting with pp = p, so don't increment before you access the array element:
left[i] = a[pp++];

Related

Merge Sort Java Implementation so merge works within one array with second splitted array reversed

Challenge: within a lecture on data structures and algorithms I encountered a version of merge sort which uses the merge routine in a way that the second half is being reversed from the splitting index and from there compares the first and the last element. I tried to implement in java and it always failed somehow.
Problem: The array is being sorted so that the output is [1, 2, 4, 8, 6] so the 6 is not sorted. It seems as if the recursive call is not looking at the element 6 in the last merge call.
What I tried: Shifting different indices and adding different print statements for checking.
I tried to make j = r before the last for loop within merge which lead to stack overflow every time. I tried to change the way how the size of the array is being calculated, since I was not sure if the pseudo code excepts the array to start from 1 or 0. I tried to shift if(p < r-1) to if(p <= r-1) but get a stack overflow.
I looked at different implementations of java merge routine and every I found so far seems to work with two arrays. Is there a serious reason why the approach above is not working correctly or any idea how to fix this issue?
Given the following pseudo code:
void merge_sort(array<T>& A, int p, int r) {
if (p < r - 1) {
int q = Floor((p + r) / 2);
merge_sort(A, p, q);
merge_sort(A, q + 1, r);
merge(A, p, q, r);
}
}
void merge(array<T>& A, int p, int q, int r) {
array<T> B(p, r - 1);
int i, j;
for (i = p; i < q; i++)
B[i] = A[i];
// Now i=q
for (j = r; i < r; i++)
B[--j] = A[i];
i = p;
j = r - 1;
for (int k = p; k < r; k++)
A[k] = (B[i] < B[j]) ? B[i++] : B[j--];
}
I tried to implement in java like so:
import java.util.Arrays;
public class Mergesort {
private static int[] A = new int[]{ 4, 2, 1, 8, 6 };
public static void main(String[] args) {
merge_sort(0, A.length - 1);
System.out.println(Arrays.toString(A));
}
public static void merge_sort(int p, int r) {
if (p < r - 1) {
int q = Math.floor((p + r) / 2);
merge_sort(p, q);
merge_sort(q + 1, r);
merge(p, q, r);
}
}
public static void merge(int p, int q, int r) {
int[] B = new int[r - p];
int i, j;
for (i = p; i < q; i++)
B[i] = A[i]
for (j = r; i < r; i++)
B[--j] = A[i];
i = p;
j = r - 1;
for (int k = p; k < r; k++)
A[k] = (B[i] < B[j])? B[i++] : B[j--];
}
}
There are multiple problems in your code:
the temporary array is too short: since r is the index of the last element, the size should be r - p + 1. It is much simpler to pass r as the index one past the last element of the slice to sort.
the first for loop is incorrect: you should use a different index into B and A.
the second for loop copies to B[r - 1] downwards, but it should use B[r - p] instead.
the merging loop is incorrect: you should test if i and j are still within the boundaries of the respective halves before accessing B[i] and/or B[j].
[minor] there is no need for int q = Math.floor((p + r) / 2); in java as p and r are have type int, so the division will use integer arithmetics.
Here is a modified version:
public class Mergesort {
private static int[] A = new int[]{ 4, 2, 1, 8, 6 };
public static void main(String[] args) {
merge_sort(0, A.length);
System.out.println(Arrays.toString(A));
}
public static void merge_sort(int p, int r) {
if (r - p >= 2) {
int q = p + (r - p) / 2;
merge_sort(p, q);
merge_sort(q, r);
merge(p, q, r);
}
}
public static void merge(int p, int q, int r) {
int m = q - p; // zero based index of the right half
int n = r - p; // length of the merged slice
int[] B = new int[n];
int i, j, k;
for (i = p, j = 0; j < m; j++)
B[j] = A[i++];
for (i = r, j = m; j < n; j++)
B[j] = A[--i];
for (i = 0, j = n, k = p; k < r; k++) {
// for stable sorting, i and j must be tested against their boundary
// A[k] = (i < m && (j <= m || B[i] <= B[j - 1])) ? B[i++] : B[--j];
// stability is not an issue for an array of int
A[k] = (B[i] <= B[j - 1]) ? B[i++] : B[--j];
}
}
}
Reversing the second half allows for a simpler merge loop without boundary tests. Note however that there is a simpler approach that uses less memory and might be more efficient:
public static void merge(int p, int q, int r) {
int m = q - p; // length of the left half
int[] B = new int[m];
int i, j, k;
// only save the left half
for (i = p, j = 0; j < m; j++)
B[j] = A[i++];
for (i = 0, j = q, k = p; i < m; k++) {
A[k] = (j >= r || B[i] <= A[j]) ? B[i++] : A[j++];
}
}

Converting Merge Sort pseudocode to running Java code

I tried to convert this Merge Sort pseudocode into Java but don't get the right output. Here is the pseudocode:
Merge-Sort(A, p, r )
if p < r
then q←(p+r)/2
Merge-Sort(A, p, q)
Merge-Sort(A, q + 1, r )
Merge(A, p, q, r )
Merge(A, p, q, r )
for k←1 to r−p+1 do
if j>r or (i ≤ q and A[i] ≤ A[j])
then B[k]←A[i]; i←i+1 else B[k]←A[j];j←j+1
for k←1 to r−p+1 do A[k+p−1]←B[k]
And this is my Java code for it:
public class MergeSort {
public static void main(String[] args) {
int[] a = {2, 6, 3, 5, 1};
mergeSort(a, 0, a.length - 1);
for (int i = 0; i < a.length; i++) {
System.out.print(" " + a[i]);
}
}
public static void mergeSort(int[] a, int from, int to) {
final int begin = from, end = to;
if (begin < end) {
final int mid = (begin + end) / 2;
MergeSort.mergeSort(a, begin, mid);
MergeSort.mergeSort(a, mid+1, end);
MergeSort.merge(a, begin, mid, end);
}
}
private static void merge(int[] a, int from, int mid, int to) {
final int begin = from, mitte = mid, end = to;
int[] B = new int[a.length];
int i = begin, j = mitte;
for (int k = 0; k <= end-begin; k++) {
if (j > end || (i <= mitte && a[i] <= a[j])) {
B[k] = a[i];
i++;
} else {
B[k] = a[j];
j++;
}
}
for (int k = 0; k < end-begin; k++) {
a[k + begin] = B[k];
}
}
Sadly it is not working like that. I think i do something wrong with some indexes but I can't figure out where exactly the error is.
I need to stick as close as possible to this pseudocode.
It would be great if someone could show me what I am doing wrong.
The pseudocode given for the Merge algorithm is somewhat incorrect because it does not say anything about the situation when only one pointer moves while other remains stationary.
In the above mentioned case you would have to separately fill out temporary array for by moving that stationary pointer.
Also the required length of B is to - from + 1 and it should be j = mitte + 1 instead of j = mitte The correct code for the merge is :
private static void merge(int[] a, int from, int mid, int to) {
final int begin = from, mitte = mid, end = to;
int[] B = new int[end-begin+1];
int k=0;
int i = begin, j = mitte+1;
while(i<=mid&&j<=end)
if(a[i]<=a[j]){
B[k++] = a[i];
i++;
} else {
B[k++] = a[j];
j++;
}
//in case i remained stationary
while(i<=mid)
B[k++] = a[i++];
//in case j remained stationary
while(j<=end)
B[k++] = a[j++];
//Now copy the array
i=0;
for(k=begin;k<=end;++k)
a[k]=B[i++];
}

What is wrong with this merge sort?

int[] input = [6,5,4,3,2,1]
mergeSort(input,0,input.length)
void mergeSort(int[] A,int p,int r){
if(p < r){
int q = (int) Math.floor((p+r)/2)
mergeSort(A,p,q)
mergeSort(A,q+1,r)
merge(A,p,q,r)
}
}
void merge(int[] A,int p,int q,int r){
int i = 0
int j = 0
int n1 = q - p + 1
int n2 = r - q
int[] L = new int[n1]
int[] R = new int[n2]
for(;i<n1;i++){ L[i] = A[p+i-1] }
for(;j<n2;j++){ R[j] = A[q+j] }
i = j = 0
for(int k = p; k < r; k++){
if(L[i] <= R[j]){ A[k] = L[i++] }
else{ A[k] = R[j++] }
}
}
This is a direct implementation of merge sort from the Introduction to Algorithms book. Although it looks correct, it ends with an ArrayIndexOutOfBounds exception.
I have been trying to debug it but couldn't. I'd like to know what's going wrong and how to correct it.
Runnable Example: http://ideone.com/GhuuSd
Firstly, although this is Groovy and therefore the negative index highlighted by #Useless won't throw an exception, it is indicative of a problem.
I think you have two problems: Caveat: I'm not familiar with the exact presentation in "Introduction to Algorithms"
Your indices into A to populate L and R are off by one - they should be [p+i] (thus avoiding the possible negative index issue) and [q+j+1]. Note with this amendment you need to pass in length-1 as the starting argument for r
in your test at the bottom to determine which of L and R to use in re-populating elements of A, you do not check for i or j running beyond the length of those arrays (i.e. is i >= n1 or j >= n2). In these cases (i.e. one of L or R has "run out" of members), you should use the other array.
Code below works for your example, I haven't tested it extensively but have tried cases with repeated numbers, negative etc and believe it should hold up:
int[] input = [6,5,4,3,2,1]
mergeSort(input,0,input.length-1)
System.out.println(Arrays.toString(input))
void mergeSort(int[] A,int p,int r){
if(p < r){
int q = (int) Math.floor((p+r)/2)
mergeSort(A,p,q)
mergeSort(A,q+1,r)
merge(A,p,q,r)
}
}
void merge(int[] A,int p,int q,int r) {
int i = 0
int j = 0
int n1 = q - p + 1
int n2 = r - q
int[] L = new int[n1]
int[] R = new int[n2]
for(;i<n1;i++){ L[i] = A[p+i] }
for(;j<n2;j++){ R[j] = A[q+j+1] }
i = j = 0
for(int k = p; k <= r; k++){
if(j >= n2 || (i < n1 && L[i] < R[j] )) { A[k] = L[i++] }
else{ A[k] = R[j++] }
}
}
int i = 0
...
for(;i<n1;i++){ L[i] = A[p+i-1] }
What is the index into A on your first iteration, where p is zero?

Merge Sort. Error-- Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 2

Good day, everyone! I have here a program that sorts 50,000 words from a file using merge sort. I followed Thomas Cormen's pseudocode in his Introduction to Algorithms and it seems right when I'm "debuuging" it by hand manually. However, when I run the program it says Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 2 . Yes, I think it is due to the large NO_OF_WORDS (ie, 50,000) but even though I decreased it to 10, still, it shows the same error.
import java.io.*;
import java.util.*;
public class SortingAnalysis {
public static void merge(String[] A, int p, int q, int r) {
int n1 = q-p+1;
int n2 = r-q;
String[] L = new String[n1+1];
String[] R = new String[n2+1];
for (int i=1; i<n1; i++) {
L[i] = A[p+i-1];
}
for (int j=1; j<n2; j++) {
R[j] = A[q+j];
}
L[n1+1] = "zzzzz"; //for infinity because if I use Math.floor, it will return a double
R[n2+1] = "zzzzz";
int i=1;
int j=1;
for (int k=p; k<=r; k++) {
int comparison = L[i].compareTo(R[j]);
if (comparison <= 0){
A[k] = L[i];
i++;
}
else {
A[k] = R[j];
j++;
}
}
}
public static void mergeSort (String[] A, int p, int r) {
if (p<r) {
int q = (p+r)/2;
mergeSort(A, p, q);
mergeSort(A, q+1, r);
merge(A, p, q, r);
}
}
public static void main(String[] args) {
final int NO_OF_WORDS = 50000;
try {
Scanner file = new Scanner(new File(args[0]));
String[] words = new String[NO_OF_WORDS];
int i = 0;
while(file.hasNext() && i < NO_OF_WORDS) {
words[i] = file.next();
i++;
}
long start = System.currentTimeMillis();
mergeSort(words, 0, words.length-1);
long end = System.currentTimeMillis();
System.out.println("Sorted Words: ");
for(int j = 0; j < words.length; j++) {
System.out.println(words[j]);
}
System.out.print("Running time: " + (end - start) + "ms");
}
catch(SecurityException securityException) {
System.err.println("Error");
System.exit(1);
}
catch(FileNotFoundException fileNotFoundException) {
System.err.println("Error");
System.exit(1);
}
}
}
I think it's because of the declaration of String[] L and R. Or not. Please help me what's the problem. Thank you very much!
EDIT
Cormen's Pseudocode
MERGE(A, p, q, r )
n1 ← q − p + 1
n2 ←r − q
create arrays L[1 . . n1 + 1] and R[1 . . n2 + 1]
for i ← 1 to n1
do L[i ] ← A[p + i − 1]
for j ← 1 to n2
do R[ j ] ← A[q + j ]
L[n1 + 1]←∞
R[n2 + 1]←∞
i ← 1
j ← 1
for k ← p to r
do if L[i ] ≤ R[ j ]
then A[k] ← L[i ]
i ←i + 1
else A[k] ← R[ j ]
j ← j + 1
I don't know what is your pseudocode but your implementation seems wrong. I've look at the wikipedia merge sort and it's quite different.
So I will not give you the full working algorithm here. I'll just give you the solution to resolve your problem of indexOutOfBounds but you still have to work more on your implementation.
In Java when you do that :
String[] L = new String[5];
You declare an array of string that can contains 5 strings within.
The access to those strings is made this way : L[anIndex].
The first element is at index 0.
So if you have an array of size 5 then the last element is at index 4 (because we start from 0).
In your code you do this :
String[] L = new String[n1+1];
String[] R = new String[n2+1];
then :
L[n1+1] = "zzzzz";
R[n2+1] = "zzzzz";
So here you always try to access a string at an index that doesn't exist.
The last element in each array is respectively n1 and n2 (because arrays size are n1+1 and n2+1 ).
I hope you'll understand better how array works in Java with this explanation. Now you have to improve your implementation because it's still not working. Maybe give us the pseudocode you use if you don't understand it well.
EDIT :
Ok I made some correction.
Here is the working algorithm. I've had to change several index to fit Java "based-0 arrays", take a look :
import java.io.*;
import java.util.*;
public class SortingAnalysis {
public static void merge(String[] A, int p, int q, int r) {
int n1 = q-p+1;
int n2 = r-q;
if(A[p]==null || A[q]==null)return;
String[] L = new String[n1+1];
String[] R = new String[n2+1];
for (int i=0; i<n1; i++) {
L[i] = A[p+i];
}
for (int j=0; j<n2; j++) {
R[j] = A[q+j +1];
}
L[n1] = "zzzzz"; //for infinity because if I use Math.floor, it will return a double
R[n2] = "zzzzz";
int i=0;
int j=0;
for (int k=p; k<=r; k++) {
int comparison = L[i].compareTo(R[j]);
if (comparison <= 0){
A[k] = L[i];
i++;
}
else {
A[k] = R[j];
j++;
}
}
}
public static void mergeSort (String[] A, int p, int r) {
if (p<r) {
int q = (p+r)/2;
mergeSort(A, p, q);
mergeSort(A, q+1, r);
merge(A, p, q, r);
}
}
public static void main(String[] args) {
final int NO_OF_WORDS = 50000;
try {
Scanner file = new Scanner("bla blya blay byla ybla");
ArrayList<String> words = new ArrayList<String>();
while(file.hasNext() && words.size() < NO_OF_WORDS) {
words.add(file.next());
}
String [] wordsArray = new String[words.size()];
words.toArray(wordsArray);
long start = System.currentTimeMillis();
mergeSort(wordsArray, 0, wordsArray.length-1);
long end = System.currentTimeMillis();
System.out.println("Sorted Words: ");
for(int j = 0; j < wordsArray.length; j++) {
System.out.println(wordsArray[j]);
}
System.out.print("Running time: " + (end - start) + "ms");
}
catch(SecurityException securityException) {
System.err.println("Error");
System.exit(1);
}
}
}
Note that I've change your Main, now I use an arrayList to avoid null value, if your text contains less words than the original array size. With your solution if you don't fill the 50000 words you get null in the array and then nullPointerException in the merge algo.
There is a big problem with your merge() method:
String[] L = new String[n1+1];
String[] R = new String[n2+1];
will not play well with
L[n1+1] = "zzzzz"; //for infinity because if I use Math.floor, it will return a double
R[n2+1] = "zzzzz";
You will get an ArrayIndexOutOfBoundsException here regardless of the values of n1 and n2 since arrays are 0-based in Java.

Merge Sort Java

I am trying to make a merge sort method, but it keeps on giving the wrong sorts. Where do I have change to make it actually sort the array? What part of the code has to be different? Thank you for your time.
public static void mergeSort(int[] array, int left, int lHigh, int right, int rHigh) {
int elements = (rHigh - lHigh +1) ;
int[] temp = new int[elements];
int num = left;
while ((left <= lHigh) && (right <= rHigh)){
if (a[left] <= array[right]) {
temp[num] = array[left];
left++;
}
else {
temp[num] = array[right];
right++;
}
num++;
}
while (left <= right){
temp[num] = array[left]; // I'm getting an exception here, and is it because of the num???
left += 1;
num += 1;
}
while (right <= rHigh) {
temp[num] = array[right];
right += 1;
num += 1;
}
for (int i=0; i < elements; i++){
array[rHigh] = temp[rHigh];
rHigh -= 1;
}
EDIT: now the mergeSort doesn't really sort the numbers, can someone tell me where it specifically is? especially when I print the "Testing merge sort" part.
First of all, I'm assuming this is academic rather than practical, since you're not using a built in sort function. That being said, here's some help to get you moving in the right direction:
Usually, one can think of a merge sort as two different methods: a merge() function that merges two sorted lists into one sorted list, and mergeSort() which recursively breaks the list into single element lists. Since a single element list is sorted already, you then merge all the lists together into one big sorted list.
Here's some off-hand pseudo-code:
merge(A, B):
C = empty list
While A and B are not empty:
If the first element of A is smaller than the first element of B:
Remove first element of A.
Add it to the end of C.
Otherwise:
Remove first element of B.
Add it to the end of C.
If A or B still contains elements, add them to the end of C.
mergeSort(A):
if length of A is 1:
return A
Split A into two lists, L and R.
Q = merge(mergeSort(L), mergeSort(R))
return Q
Maybe that'll help clear up where you want to go.
If not, there's always MergeSort at wikipedia.
Additional:
To help you out, here are some comments inline in your code.
public static void mergeSort(int[] array, int left, int lHigh, int right, int rHigh) {
// what do lHigh and rHigh represent?
int elements = (rHigh - lHigh +1) ;
int[] temp = new int[elements];
int num = left;
// what does this while loop do **conceptually**?
while ((left <= lHigh) && (right <= rHigh)){
if (a[left] <= a[right]) {
// where is 'pos' declared or defined?
temp[pos] = a[left];
// where is leftLow declared or defined? Did you mean 'left' instead?
leftLow ++;
}
else {
temp[num] = a[right];
right ++;
}
num++;
}
// what does this while loop do **conceptually**?
while (left <= right){
// At this point, what is the value of 'num'?
temp[num] = a[left];
left += 1;
num += 1;
}
while (right <= rHigh) {
temp[num] = a[right];
right += 1;
num += 1;
}
// Maybe you meant a[i] = temp[i]?
for (int i=0; i < elements; i++){
// what happens if rHigh is less than elements at this point? Could
// rHigh ever become negative? This would be a runtime error if it did
a[rHigh] = temp[rHigh];
rHigh -= 1;
}
I'm purposefully being vague so you think about the algorithm. Try inserting your own comments into the code. If you can write what is conceptually happening, then you may not need Stack Overflow :)
My thoughts here are that you are not implementing this correctly. This is because it looks like you're only touching the elements of the array only once (or close to only once). This means you have a worst case scenario of O(N) Sorting generally takes at least O(N * log N) and from what I know, the simpler versions of merge sort are actually O(N^2).
More:
In the most simplistic implementation of merge sort, I would expect to see some sort of recursion in the mergeSort() method. This is because merge sort is generally defined recursively. There are ways to do this iteratively using for and while loops, but I definitely don't recommend it as a learning tool until you get it recursively.
Honestly, I suggest taking either my pseudo-code or the pseudo-code you may find in a wikipedia article to implement this and start over with your code. If you do that and it doesn't work correctly still, post it here and we'll help you work out the kinks.
Cheers!
And finally:
// Precondition: array[left..lHigh] is sorted and array[right...rHigh] is sorted.
// Postcondition: array[left..rHigh] contains the same elements of the above parts, sorted.
public static void mergeSort(int[] array, int left, int lHigh, int right, int rHigh) {
// temp[] needs to be as large as the number of elements you're sorting (not half!)
//int elements = (rHigh - lHigh +1) ;
int elements = rHigh - left;
int[] temp = new int[elements];
// this is your index into the temp array
int num = left;
// now you need to create indices into your two lists
int iL = left;
int iR = right;
// Pseudo code... when you code this, make use of iR, iL, and num!
while( temp is not full ) {
if( left side is all used up ) {
copy rest of right side in.
make sure that at the end of this temp is full so the
while loop quits.
}
else if ( right side is all used up) {
copy rest of left side in.
make sure that at the end of this temp is full so the
while loop quits.
}
else if (array[iL] < array[iR]) { ... }
else if (array[iL] >= array[iR]) { ... }
}
}
public class MergeSort {
public static void main(String[] args) {
int[] arr = {5, 4, 7, 2, 3, 1, 6, 2};
print(arr);
new MergeSort().sort(arr, 0, arr.length - 1);
}
private void sort(int[] arr, int lo, int hi) {
if (lo < hi) {
int mid = (lo + hi) / 2;
sort(arr, lo, mid); // recursive call to divide the sub-list
sort(arr, mid + 1, hi); // recursive call to divide the sub-list
merge(arr, lo, mid, hi); // merge the sorted sub-lists.
print(arr);
}
}
private void merge(int[] arr, int lo, int mid, int hi) {
// allocate enough space so that the extra 'sentinel' value
// can be added. Each of the 'left' and 'right' sub-lists are pre-sorted.
// This function only merges them into a sorted list.
int[] left = new int[(mid - lo) + 2];
int[] right = new int[hi - mid + 1];
// create the left and right sub-list for merging into original list.
System.arraycopy(arr, lo, left, 0, left.length - 1);
System.arraycopy(arr, mid + 1, right, 0, left.length - 1);
// giving a sentinal value to marking the end of the sub-list.
// Note: The list to be sorted is assumed to contain numbers less than 100.
left[left.length - 1] = 100;
right[right.length - 1] = 100;
int i = 0;
int j = 0;
// loop to merge the sorted sequence from the 2 sub-lists(left and right)
// into the main list.
for (; lo <= hi; lo++) {
if (left[i] <= right[j]) {
arr[lo] = left[i];
i++;
} else {
arr[lo] = right[j];
j++;
}
}
}
// print the array to console.
private static void print(int[] arr) {
System.out.println();
for (int i : arr) {
System.out.print(i + ", ");
}
}
}
Here's another!
private static int[] mergeSort(int[] input){
if (input.length == 1)
return input;
int length = input.length/2;
int[] left = new int[length];
int[] right = new int[input.length - length];
for (int i = 0; i < length; i++)
left[i] = input[i];
for (int i = length; i < input.length; i++)
right[i-length] = input[i];
return merge(mergeSort(left),mergeSort(right));
}
private static int[] merge(int[] left, int[] right){
int[] merged = new int[left.length+right.length];
int lengthLeft = left.length;
int lengthRight = right.length;
while (lengthLeft > 0 && lengthRight > 0){
if (left[left.length - lengthLeft] < right[right.length - lengthRight]){
merged[merged.length -lengthLeft-lengthRight] = left[left.length - lengthLeft];
lengthLeft--;
}else{
merged[merged.length - lengthLeft-lengthRight] = right[right.length - lengthRight];
lengthRight--;
}
}
while (lengthLeft > 0){
merged[merged.length - lengthLeft] = left[left.length-lengthLeft];
lengthLeft--;
}
while (lengthRight > 0){
merged[merged.length - lengthRight] = right[right.length-lengthRight];
lengthRight--;
}
return merged;
}
static void mergeSort(int arr[],int p, int r) {
if(p<r) {
System.out.println("Pass "+k++);
int q = (p+r)/2;
mergeSort(arr,p,q);
mergeSort(arr,q+1,r);
//System.out.println(p+" "+q+" "+r);
merge(arr,p,q,r);
}
}
static void merge(int arr[],int p,int q,int r) {
int temp1[],temp2[];
//lower limit array
temp1 = new int[q-p+1];
//upper limit array
temp2 = new int[r-q];
for(int i=0 ; i< (q-p+1); i++){
temp1[i] = arr[p+i];
}
for(int j=0; j< (r-q); j++){
temp2[j] = arr[q+j+1];
}
int i = 0,j=0;
for(int k=p;k<=r;k++){
// This logic eliminates the so called sentinel card logic mentioned in Coreman
if(i!= temp1.length
&& (j==temp2.length || temp1[i] < temp2[j])
) {
arr[k] = temp1[i];
// System.out.println(temp1[i]);
i++;
}
else {
//System.out.println(temp2[j]);
arr[k] = temp2[j];
j++;
}
}
}
>
Merge Sort Using Sentinel
This codes works perfectly fine.
public void mergeSort(int a[], int low, int high) {
if (low < high) {
int mid = (low + high) / 2;
mergeSort(a, low, mid);
mergeSort(a, mid + 1, high);
merge(a, low, mid, high);
}
}
public void merge(int a[], int low, int mid, int high) {
int n1 = mid - low + 1;// length of an array a1
int n2 = high - mid; // length of an array a2
int a1[] = new int[n1 + 1];
int a2[] = new int[n2 + 1];
int lowRange = low;
for (int i = 0; i < n1; i++) {
a1[i] = a[lowRange];
lowRange++;
}
for (int j = 0; j < n2; j++) {
a2[j] = a[mid + j + 1];
}
a1[n1] = Integer.MAX_VALUE; // inserting sentinel at the end of array a1
a2[n2] = Integer.MAX_VALUE; // inserting sentinel at the end of array a2
int i = 0;
int j = 0;
int k = low;
for (k = low; k <= high; k++) {
if (a1[i] >= a2[j]) {
a[k] = a2[j];
j++;
} else {
a[k] = a1[i];
i++;
}
}
if (a2.length >= a1.length) {
for (int ab = k; ab < a2.length; ab++) {
a[k] = a2[ab];
k++;
}
} else if (a1.length >= a2.length) {
for (int ab = k; ab < a1.length; ab++) {
a[k] = a1[ab];
k++;
}
}
}
Here's another alternative:
public class MergeSort {
public static void merge(int[]a,int[] aux, int f, int m, int l) {
for (int k = f; k <= l; k++) {
aux[k] = a[k];
}
int i = f, j = m+1;
for (int k = f; k <= l; k++) {
if(i>m) a[k]=aux[j++];
else if (j>l) a[k]=aux[i++];
else if(aux[j] > aux[i]) a[k]=aux[j++];
else a[k]=aux[i++];
}
}
public static void sort(int[]a,int[] aux, int f, int l) {
if (l<=f) return;
int m = f + (l-f)/2;
sort(a, aux, f, m);
sort(a, aux, m+1, l);
merge(a, aux, f, m, l);
}
public static int[] sort(int[]a) {
int[] aux = new int[a.length];
sort(a, aux, 0, a.length-1);
return a;
}
}
Here is a simple merge sort algorithm in Java:
Good Tip: Always use int middle = low + (high-low)/2 instead of int middle = (low + high)/2.
public static int[] mergesort(int[] arr) {
int lowindex = 0;
int highindex = arr.length-1;
mergesort(arr, lowindex, highindex);
return arr;
}
private static void mergesort(int[] arr, int low, int high) {
if (low == high) {
return;
} else {
int midIndex = low + (high-low)/2;
mergesort(arr, low, midIndex);
mergesort(arr, midIndex + 1, high);
merge(arr, low, midIndex, high);
}
}
private static void merge(int[] arr, int low, int mid, int high) {
int[] left = new int[mid-low+2];
for (int i = low; i <= mid; i++) {
left[i-low] = arr[i];
}
left[mid-low+1] = Integer.MAX_VALUE;
int[] right = new int[high-mid+1];
for (int i = mid+1; i <= high; i++) {
right[i-mid-1] = arr[i];
}
right[high - mid] = Integer.MAX_VALUE;
int i = 0;
int j = 0;
for (int k = low; k <= high; k++) {
if (left[i] <= right[j]) {
arr[k] = left[i];
i++;
} else {
arr[k] = right[j];
j++;
}
}
}
package com.sortalgo;
import java.util.Arrays;
public class MyMSort {
private static void merge(int[] array, int[] result, int low, int mid, int high) {
int k =low, i=low; int j=mid+1;
while(i<=mid && j<=high) {
if(array[i]<= array[j]) {
result[k++]=array[i++];
}else {
result[k++]=array[j++];
}
}
while(i<=mid) {
result[k++]=array[i++];
}
while(j<=high) {
result[k++]=array[j++];
}
for(i=low;i<=high;i++) {
array[i]=result[i];
}
}
private static void mergeSort(int[] array, int[] result, int low, int high) {
if(high == low) {
return ;
}
int mid = (low + high)/2;
mergeSort(array,result, low, mid );
mergeSort(array,result, mid+1, high );
merge(array, result, low, mid, high);
}
public static void main(String[] args) {
int[] array = {8,4,3,12,25,6,13,10};
int[] result = new int[array.length];
mergeSort(array, result, 0, array.length-1 );
for(int i=0; i<=array.length-1;i++) {
System.out.println(array[i]);
}
}
}

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