# 1046. Last Stone Weight 最后一块石头的重量

@TOC

## # 题目描述

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

## # 解题方法

### # 大根堆

Python代码如下：

``````class Solution(object):
def lastStoneWeight(self, stones):
"""
:type stones: List[int]
:rtype: int
"""
stones = map(lambda x : -x, 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]
``````

C++代码如下：

``````class Solution {
public:
int lastStoneWeight(vector<int>& stones) {
priority_queue<int> q;
for (int s : stones)
q.push(s);
while (q.size() >= 2) {
int a = q.top(); q.pop();
if (!q.empty()) {
int b = q.top(); q.pop();
if (a != b) {
q.push(abs(a - b));
}
}
}
return q.empty() ? 0 : q.top();
}
};
``````

## # 日期

2019 年 6 月 8 日 —— 刷题尽量不要停