Formatted question description: https://leetcode.ca/all/1046.html

# 1046. Last Stone Weight (Easy)

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  then that's the value of last stone.

Note:

1. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

Related Topics:
Heap, Greedy

## Solution 1. Max-root Heap

// OJ: https://leetcode.com/problems/last-stone-weight/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int lastStoneWeight(vector<int>& A) {
priority_queue<int> pq(begin(A), end(A));
while (pq.size() > 1) {
int a = pq.top();
pq.pop();
int b = pq.top();
pq.pop();
pq.push(a - b);
}
return pq.top();
}
};


Java

class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(new Comparator<Integer>() {
public int compare(Integer num1, Integer num2) {
return num1 == null || num2 == null ? 0 : num2 - num1;
}
});
for (int stone : stones)
priorityQueue.offer(stone);
while (priorityQueue.size() > 1) {
int stone1 = priorityQueue.poll();
int stone2 = priorityQueue.poll();
if (stone1 > stone2)
priorityQueue.offer(stone1 - stone2);
}
return priorityQueue.size() == 0 ? 0 : priorityQueue.poll();
}
}