4月30天leetcode训练-Day12

Day12-Last Stone Weight

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.)

这道题描述的应该就是数据结构中的大根堆的情况吧，每次从堆中拿出两个最大的相减，再放回堆中（0不放回去），直到堆中只有一个元素或者没有元素为止。

Example:

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.

解法一：

我们遍历这个数组中元素，每遍历一个元素之前先排序，然后排出两个最大的，再把最大的加进去，最后返回第一个元素就行了。

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        for i in range(len(stones) - 1):
            stones.sort()
            stones.append(stones.pop() - stones.pop())
            
        return stones[0]

时间复杂度：这种就是纯暴力，时间复杂度高的多了，python的sort函数的时间复杂度为nlogn,然后还在循环里面应该就是O(n*nlogn).这解法也就一乐。还得看下面的大根堆的解法。

解法二：

用到了python的heap数据结构，但是python的是小根堆，为此我们先取反，然后再最后返回的时候加上负号就行了。

class Solution(object):
    def lastStoneWeight(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        stones = [-x for x in stones]
        heapq.heapify(stones)
        while len(stones) > 1:
            x = heapq.heappop(stones)
            if stones:
                y = heapq.heappop(stones)
                if x != y:
                    heapq.heappush(stones, -abs(x - y))
        return 0 if not stones else -stones[0]

时间复杂度为O(nlogn).其实我觉得这两种做法都不太符合出题者的要求吧，出题者可能是要我们自己手动去实现一个大根堆的解法，而不是用python的内置函数来做。。。

有时间会再手写一个大根堆的解法。