题目
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.)
解题思路
使用优先队列，将石子从放到优先队列（大头）中，每次pop出来最大的两个，做差（如果非0）将绝对值加回队里中，直到结束
代码
import heapq
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
stones = [-s for s in stones]
heapq.heapify(stones)
while len(stones) >= 2:
s1 = heapq.heappop(stones)
s2 = heapq.heappop(stones)
if s1 != s2:
heapq.heappush(stones, s1 - s2)
return -stones[0] if stones else 0
