# Java : Sort integer array without using Arrays.sort()

## Java : Sort integer array without using Arrays.sort()

Simple sorting algorithm Bubble sort:

``````public static void main(String[] args) {
int[] arr = new int[] { 6, 8, 7, 4, 312, 78, 54, 9, 12, 100, 89, 74 };

for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
int tmp = 0;
if (arr[i] > arr[j]) {
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
}
}
}
``````

You can find so many different sorting algorithms in internet, but if you want to fix your own solution you can do following changes in your code:

`````` orderedNums[greater]=tenNums[indexL];
``````

you need to do this:

``````while (orderedNums[greater] == tenNums[indexL]) {
greater++;
}
orderedNums[greater] = tenNums[indexL];
``````

This code basically checks if that particular index is occupied by a similar number, then it will try to find next free index.

Note: Since the default value in your sorted array elements is 0, you need to make sure 0 is not in your list. otherwise you need
to initiate your sorted array with an especial number that you sure is
not in your list e.g: `Integer.MAX_VALUE`

#### Java : Sort integer array without using Arrays.sort()

Simple way :

``````int a[]={6,2,5,1};
System.out.println(Arrays.toString(a));
int temp;
for(int i=0;i<a.length-1;i++){
for(int j=0;j<a.length-1;j++){
if(a[j] > a[j+1]){   // use < for Descending order
temp = a[j+1];
a[j+1] = a[j];
a[j]=temp;
}
}
}
System.out.println(Arrays.toString(a));

Output:
[6, 2, 5, 1]
[1, 2, 5, 6]
``````