1046. 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.)

 

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

这题没啥好说的，按照流程走就可以了。

class LastStoneWeight : public Solution {
public:
    void Exec() {

    }
    int lastStoneWeight(vector<int>& stones) {
        while (stones.size() > 1) {
            std::sort(stones.begin(), stones.end());
            std::size_t sz = stones.size();
            int diff = stones[sz - 1] - stones[sz - 2];
            stones.pop_back();
            stones.pop_back();
            if (diff != 0) {
                stones.push_back(diff);
            }
        }
        if (stones.empty()) {
            return 0;
        }
        return stones.front();
    }
};