Thread: help in sorting the array

i'm trying to sort this array in special method by :
- Counting the number of smaller elements to find the correct position i.
– If the element is in its correct position, move to the succeeding element.
– Otherwise, swap the current element with the one found in position i.
– Repeat the previous steps till you reach the last element.
So i did this code but something wrong so please help :


public class Ass {

public static void swap(int [] x) {
int i = 0 ;
while (i<x.length){

for (int j = 0 ; j<x.length ; j++){
int b = 0 ;
if ( x[i] > x[j] ) {
b++ ;
}
if (b!=0) {
int tmp = x[i] ;
x[i]=x[b] ;
x[b]= tmp ;

}





}
i++ ;
}
}
public static void main (String [] args) {
int [] array = {5,7,3,6,9} ;
swap(array) ;
for (int k = 0 ; k<array.length ; k++)
System.out.print(array[k] + " ");
}
}

Please use code tags as described in the announcements page as well as in your previous thread.

Hi,
I checked your code, even executed the same and what I found was,
a change is needed in the swap code.
what you have written is :

 int tmp = x[i] ;
 x[i]=x[b] ;
 x[b]= tmp ;

which should be as follows :

 int tmp = x[i] ;
 x[i]=x[j] ;
 x[j]= tmp ;

Because, since you are comparing the values at i and j positions, and if you find value at i is greater than value at j, then you should swap the values at i and j only but right now you are performing the swap between the values at i and b, which is wrong.

b is just like a check variable ,to find out whether the value at ' i greater than j ' condition is true or false.

actually there is no need of the variable b. you can compare the values and directly swap like below :

     if (x[i] > x[j]) {
			  int tmp = x[i];
			  x[i] = x[j];
			  x[j] = tmp;
 
			}

thnkxxx so muchh but this prints the biggest first i want the smallest , any help ??

nvmmm i solved it thnkxxx

@ind
Please see The problem with spoonfeeding.

Since the OP has already taken the code provided and run with it, I am not going to delete it, but please see the link and try to be more suggestive and a little less free answers in your posts.