1046. Last Stone Weight

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

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给定一个数组，每个数表示一个石头的质量，每次从数组中选两个最终的石头，如果相等，则两个石头都砸碎，不相等这把质量差再次放入数组。问最终剩余质量是多少。

简单模拟题，使用最大堆来装所有石头的质量，代码如下：

class Solution {
public:
	int lastStoneWeight(vector<int>& stones) {
		priority_queue<int,vector<int>,std::less<int>> pq(stones.begin(),stones.end());
		while (pq.size() >= 2) {
			int first = pq.top();
			pq.pop();
			int second = pq.top();
			pq.pop();

			if (first != second)pq.push(first - second);
		}
		if (pq.size() == 0)return 0;
		else return pq.top();
	}
};

本代码提交AC，用时0MS。