Last Stone Weight

Description

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 [1] then that's the value of last stone.

Note:

1. 1 <= stones.length <= 30

2. 1 <= stones[i] <= 1000

Code

class Solution {
    public int lastStoneWeight(int[] stones) {
        Queue<Integer> queue = new PriorityQueue<>(stones.length, Comparator.reverseOrder());
        for (int s : stones) {
            queue.offer(s);
        }
        while (queue.size() > 1) {
            Integer first = queue.poll();
            Integer second = queue.poll();
            if (first - second > 0) {
                queue.add(first - second);
            }
        }
        return queue.isEmpty() ? 0 : queue.poll();
    }
}