leetcode-10460-easy

Last Stone Weight

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 weight of the last remaining 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.
Example 2:

Input: stones = [1]
Output: 1
Constraints:

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

思路一：有优先队列存储，模拟操作

    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> queue = new PriorityQueue<>(Comparator.reverseOrder());

        for (int stone : stones) {
            queue.offer(stone);
        }

        while (queue.size() > 1) {
            Integer x = queue.poll();
            Integer y = queue.poll();

            int abs = Math.abs(x - y);
            if (abs > 0) {
                queue.offer(abs);
            }
        }

        return queue.isEmpty() ? 0 : queue.poll();
    }