1046 Last Stone Weight

1 题目

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

2 尝试解

2.1 分析

给定一组数组，每个数代表一块石头的重量。每次挑出其中最大的两块相撞，如果两块石头大小相等，那么都消失，否则剩余一块石头，质量为二者之差。问最后剩余的石头质量。

用一个堆存储石头，每次取前两块，进行处理，然后将结果重新放入堆中，直到其中只剩一块石头。

2.2 代码

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue<int> saver;
        for(auto stone : stones){
            saver.push(stone);
        }
        while(!saver.empty()){
            int first = saver.top();
            saver.pop();
            if(saver.empty())
                return first;
            else{
                int second = saver.top();
                saver.pop();
                first = abs(second - first);
                saver.push(first);
            }
        }
        return 0;
    }
};

3 标准解

class Solution {
public:
    int lastStoneWeight(vector<int>& st) {
      multiset<int> s(begin(st), end(st));
      while (s.size() > 1) {
        auto w1 = *prev(s.end());
        s.erase(prev(s.end()));
        auto w2 = *prev(s.end());
        s.erase(prev(s.end()));
        if (abs(w1 - w2) > 0) s.insert(abs(w1 - w2));
      }
      return s.empty() ? 0 : *s.begin();
    }
};