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.

Note:

1. 1 <= stones.length <= 30

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

int lastStoneWeight(vector<int>& stones) { // time: O(nlogn); space: O(n)
    priority_queue<int> pq(stones.begin(), stones.end());
    int x, y;
    while (pq.size() > 1) {
        y = pq.top(); pq.pop();
        x = pq.top(); pq.pop();
        if (y > x) pq.push(y - x);
    }
    return pq.empty() ? 0 : pq.top();
}