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Wednesday, February 26, 2014

Quick sort in java

Input must be a list of distinct integers. Time complexity is O(n2
import java.io.*;
import java.util.*;

public class QuickSort
{
public static void swap (int A[], int x, int y)
{
int temp = A[x];
A[x] = A[y];
A[y] = temp;
}

// Reorganizes the given list so all elements less than the first are
// before it and all greater elements are after it.
public static int partition(int A[], int f, int l)
{
int pivot = A[f];
while (f < l)
{
if (A[f] == pivot || A[l] == pivot)
{
System.out.println("Only distinct integers allowed - C321");
System.out.println("students should ignore this if statement");
System.out.exit(0);
}
while (A[f] < pivot) f++;
while (A[l] > pivot) l--;
swap (A, f, l);
}
return f;
}

public static void Quicksort(int A[], int f, int l)
{
if (f >= l) return;
int pivot_index = partition(A, f, l);
Quicksort(A, f, pivot_index);
Quicksort(A, pivot_index+1, l);
}

// Usage: java QuickSort [integer] ...
// All integers must be distinct
public static void main(String argv[])
{
int A[] = new int[argv.length];
for (int i=0 ; i < argv.length ; i++)
A[i] = Integer.parseInt(argv[i]);

Quicksort(A, 0, argv.length-1);

for (int i=0 ; i < argv.length ; i++) System.out.print(A[i] + " ");
System.out.println();
}
}

2 comments :

for IT the said...

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Kanye Co Jamila said...

Great Article
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