1046 Last Stone Weight

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

If x == y, both stones are totally destroyed;
If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

 

Example 1:

Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

public int lastStoneWeight(int[] stones) {
      PriorityQueue<Integer> maxHeap = new PriorityQueue(Collections.reverseOrder());
        
     for(int i=0;i<stones.length;i++){
         maxHeap.add(stones[i]);
     }
        
    while(maxHeap.size()>1){
        
        int last = maxHeap.poll(); 
        int last2 = maxHeap.poll();
        if(last!=last2){
            int newStone = last-last2;
            maxHeap.add(newStone); 
        }
    }
    
    if(maxHeap.isEmpty())return 0;
    else return maxHeap.poll();   
    }