1046. 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.

Note:

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

class Solution {
    public int lastStoneWeight(int[] A) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b)-> b - a);
        for (int a : A)
            pq.offer(a);
        while (pq.size() > 1)
            pq.offer(pq.poll() - pq.poll());
        return pq.poll();
    }

}

https://leetcode.com/problems/last-stone-weight/discuss/294956/JavaC%2B%2BPython-Priority-Queue

大神总是能一次次的使我下巴脱臼

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12245054.html