leetcode 1046 - Last Stone Weight

https://leetcode.com/problems/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 <= stones.length <= 30
1 <= stones[i] <= 1000

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

for (int s : stones)
pq.push(s);

while (pq.size() >= 2) {
int s1 = pq.top(); pq.pop();
int s2 = pq.top(); pq.pop();

int s = abs(s1 - s2);
if (s)
pq.push(s);
}

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