 ### LeetCode - Arrays - 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  then that's the value of last stone.```
```class Solution {     public int lastStoneWeight(int[] stones) {         List<Integer> stone_list = Arrays.stream(stones).boxed().collect(Collectors.toList());         Collections.sort(stone_list);         int size = stone_list.size();         while(true){             if(size == 0) return 0;             else if(size == 1) return stone_list.get(0);             else{                 int x = stone_list.get(size-1);                 int y = stone_list.get(size-2);                 stone_list.remove(size-1);                 stone_list.remove(size-2);                 if(x != y){                     stone_list.add(x-y);                 }                 Collections.sort(stone_list);                 size = stone_list.size();             }         }     } }```