Leetcode 1046

Very straightforward question, just mimc the instruction using priority queue.

class Solution {
    public int lastStoneWeight(int[] stones) {
        // very straight forward, we use priority queue to ensure we polling the largest 2 stones
        PriorityQueue<Integer> pq = new PriorityQueue<>((a,b)->{return b - a;});
        for(int stone : stones){
            pq.add(stone);
        }
        // while we have 2 or more stones
        while(pq.size() > 1){
            int a = pq.poll();
            int b = pq.poll();
            if(a != b){
                int remain = Math.abs(a - b);
                pq.add(remain);
            }
        }
        // return nothing if we got no stone or last stone standing.
        return pq.isEmpty() ? 0 : pq.poll();
    }
}