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moni sharma
quick sort
1 Answer(s)      3 years and 8 months ago
Posted in : Java Beginners


sir, i try to modify this one as u sugess me in previous answer "array based problem" for run time input.but i am facing some problem.plz solve this one also.

import java.util.*; public class QuickSort1 { public static void quick_srt(int array[],int low, int n){ int lo = low; int hi = n; if (lo >= n) { return; } int mid = array[(lo + hi) / 2]; while (lo < hi) { while (lo mid) { hi--; } if (lo < hi) { int T = array[lo]; array[lo] = array[hi]; array[hi] = T; } } if (hi < lo) { int T = hi; hi = lo; lo = T; } quick_srt(array, low, lo); quick_srt(array, lo == low ? lo+1 : lo, n); }

public static void main(String a[]){ Scanner input=new Scanner(System.in); int array[]=new int[10]; int i;

    System.out.println("       Quick Sort\n\n");   
    System.out.println("Values Before the sort:\n");    
    for(i = 0; i < array.length; i++){
    array[i]=input.nextInt();
     }
    for(i = 0; i < array.length; i++){
    System.out.println(array[i]);
        }



  quick_srt(array,0,array.length-1);
  System.out.print("Values after the sort:\n");    
  for(i = 0; i <array.length; i++)
  System.out.print(array[i]+"  ");
   System.out.println();  

  System.out.println("PAUSE");

}

}


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January 5, 2011 at 10:14 AM


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