1046.Last Stone Weight

Tags: Easy Heap

Links: https://leetcode.com/problems/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.

---

Answer:

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue<int> pq;

        for (int i = 0; i < stones.size(); ++i){
            pq.push(stones[i]);
        }

        while (pq.size() >= 2){
            int x = pq.top();
            pq.pop();
            int y = pq.top();
            pq.pop();
            if (x != y)
                pq.push(x - y);
        }

        return pq.size() == 1 ? pq.top() : 0;
    }
};