1046. Last Stone Weight --------------E

Q:

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

A:

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        while(stones.size()>1){
            sort(stones.begin(), stones.end());
            int y = stones.back();
            stones.pop_back();
            int x = stones.back();
            stones.pop_back();
            if(x<y){
                int diff = y - x;
                stones.push_back(diff);
            }
        }
        if(stones.size()==0)
            return 0;
        return stones[0];
    }
};