原题链接在这里：https://leetcode.com/problems/last-stone-weight/
题目：
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 weightxis totally destroyed, and the stone of weightyhas new weighty-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 <= 301 <= stones[i] <= 1000
题解：
Put all stones into max heap.
While heap size >= 2, poll top 2 elements and get diff. If diff is larger than 0, add it back to heap.
Time Complexity: O(nlogn). n = stones.length. while loop could run for maximumn n-1 times.
Space: O(n).
AC Java:
1 class Solution {
2 public int lastStoneWeight(int[] stones) {
3 if(stones == null || stones.length == 0){
4 return 0;
5 }
6
7 PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
8 for(int stone : stones){
9 maxHeap.add(stone);
10 }
11
12 while(maxHeap.size() > 1){
13 int x = maxHeap.poll();
14 int y = maxHeap.poll();
15 int diff = x-y;
16 if(diff > 0){
17 maxHeap.add(diff);
18 }
19 }
20
21 return maxHeap.isEmpty() ? 0 : maxHeap.peek();
22 }
23 }