题目标签：Greedy
利用priority queue， 把石头重量都存入 pq， 每次取最大两个比较，存入差值，直到pq 只剩最后一个。
Java Solution:
Runtime: 1 ms, faster than 92.5%
Memory Usage: 37.1 MB, less than 100%
完成日期：02/14/2020
关键点：priority queue
class Solution { public int lastStoneWeight(int[] stones) { PriorityQueue <Integer> pq = new PriorityQueue<>((a, b) -> b - a); // add each stone weight into pq for(int s : stones) { pq.offer(s); } // each time get two max stones and add the difference back into pq while(pq.size() > 1) { pq.offer(pq.poll() - pq.poll()); } return pq.poll(); } }
参考资料：LeetCode Discuss
LeetCode 题目列表 - LeetCode Questions List
题目来源：https://leetcode.com/
LeetCode 1046. Last Stone Weight (最后一块石头的重量 )
原文：https://www.cnblogs.com/jimmycheng/p/12310516.html