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.)

Input:[2,7,4,1,8,1]Output:1Explanation: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.

- 1 <= stones.length <= 30
- 1 <= stones[i] <= 1000

Solution:

class Solution { public int lastStoneWeight(int[] stones) { PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a); for (int a : stones) pq.offer(a); while (pq.size() > 1) pq.offer(pq.poll() - pq.poll()); return pq.poll(); } }