leetcode 1046. Last Stone Weight（最后石头的重量）

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are destroyed, and
If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.
At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

Example 1:

Input: stones = [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 the last stone.

数组stones，里面数字是石头的重量。
每次取出最重的两颗石头，如果重量相等就毁掉，重量不相等，则剩下重量为差绝对值的石头。
返回剩下的最后一颗石头的重量，如果不剩下，返回0。

思路：
既然每次取出最重的两颗，显然是要排序的，
但是得到了差的绝对值之后，放回去还要再排序，

比如[2,7,4,1,8,1]，排序成[1, 1, 2, 4, 7, 8]
由7，8得到1后，把1放回去，又要排序，这显然很耗时，所以用最大堆，保证每次取的都是最大的元素。

最后可能都消掉了，堆里没有元素，这时返回0，
堆里剩一个的时候返回剩下的元素。

    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> (b - a));
        
        for(int stone : stones) heap.add(stone);
        
        while(heap.size() > 1) {
            int stone1 = heap.poll();
            int stone2 = heap.poll();
            
            if(stone1 == stone2) continue;  
             
            heap.add(stone1 - stone2);
        }
        return heap.size() == 0 ? 0 : heap.poll();
    }