Description

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 [1] then that's the value of last stone.

Note:

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

分析

题目的意思是：给定stones数组，按照上面的规则（x=y的时候，两数合并为0，x!=y的时候，两数合并为y-x），每次合并两个数，问最后剩下的数是多少。

思路就是最大堆，每次取出两个最大的数，计算后存回堆中，进行调整后，又取出两个最大的数，知道堆中只有一个数为止。

由于python的堆是最小堆，我这里全部变成相反数，这样取出来的数就是最大堆了，哈哈。

代码

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        stones=[-item for item in stones]
        heapq.heapify(stones)
        while(len(stones)>1):
            cur1=heapq.heappop(stones)
            cur2=heapq.heappop(stones)
            remain=cur1-cur2
            heapq.heappush(stones,remain)
        return -heapq.heappop(stones)

参考文献

​​Python的heapq模块实现大顶堆，小顶堆​​