Optimized Bubble Sort (Java)

First of all, you have an out-of-bounds access:

    for(int j=0; j<a.length; j++) {
      if(a[j] > a[j+1]) {

for j == a.length-1, so the loop condition should rather be j < a.length-1.

But, in Bubble sort, you know that after k passes, the largest k elements are sorted at the k last entries of the array, so the conventional Bubble sort uses

public static void bubblesort(int[] a) {
  for(int i=1; i<a.length; i++) {
    boolean is_sorted = true;

    for(int j=0; j < a.length - i; j++) { // skip the already sorted largest elements
      if(a[j] > a[j+1]) {
         int temp = a[j];
         a[j] = a[j+1];
         a[j+1] = temp;
         is_sorted = false;
      }
    }

    if(is_sorted) return;
  }
}

Now, that would still do a lot of unnecessary iterations when the array has a long sorted tail of largest elements, say you have k,k-1,...,1 as the first k elements and k+1 to 100000000 in order after that. The standard Bubble sort will pass k times through (almost) the entire array.

But if you remember where you made your last swap, you know that after that index, there are the largest elements in order, so

public static void bubblesort(int[] a) {
  int lastSwap = a.length-1;
  for(int i=1; i<a.length; i++) {
    boolean is_sorted = true;
    int currentSwap = -1;

    for(int j=0; j < lastSwap; j++) {
      if(a[j] > a[j+1]) {
         int temp = a[j];
         a[j] = a[j+1];
         a[j+1] = temp;
         is_sorted = false;
         currentSwap = j;
      }
    }

    if(is_sorted) return;
    lastSwap = currentSwap;
  }
}

would sort the above example with only one pass through the entire array, and the remaining passes only through a (short) prefix.

Of course, in general, that won't buy you much, but then optimising a Bubble sort is a rather futile exercise anyway.

you should use a variable "size" for the inner loop and change it to the latest swapped element in each cycle.This way your inner loop goes up to the latest "swapped" element and passes the rest that are unswapped (aka in their correctplace). i.e

do {
        int newsize =0;
        for (int i = 1; i < size; i++) {
            if (a[i - 1] > a[i]) {
                int temp;
                temp = a[i - 1];
                a[i - 1] = a[i];
                a[i] = temp;
                newsize =i;
            }
        }
        size = newsize;
   } while (size > 0);