1046. Last Stone Weight

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two heaviest rocks 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.

使用大堆来存这些石头，不断从堆里拿出两个来，比较。得到的差值如果不等于0就再放入堆里。
直到堆的大小小于2.

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(
                                            (o1, o2) -> Integer.compare(o2, o1));
        for (int stone : stones) {
            maxHeap.add(stone);
        }
        while (maxHeap.size() >= 2) {
            int x = maxHeap.poll();
            int y = maxHeap.poll();
            int z = x - y;
            if (z != 0) {
                maxHeap.add(z);
            }
        }
        return maxHeap.size() == 0 ? 0 : maxHeap.poll();
    }
}

来源：https://blog.csdn.net/ALLENYYE/article/details/102778339