Last Stone Weight.cpp

/*

 Last Stone Weight
Solution
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

*/

class Solution {
public:
    int res=0;
    public:
    int lastStoneWeight(vector<int>& stones) {
        int res = 0;
        while(stones.size()>1)
        {
            sort(stones.begin(),stones.end(),greater<int>());
            cout<<stones.size();
            if(stones[0]>stones[1])
            {
                stones[0] = stones[0] - stones[1];
                stones.erase(stones.begin()+1);
            }else{
                stones.erase(stones.begin(),stones.begin()+2);
            }
            cout<<"  ->  "<<stones.size()<<endl;
        }
        if(stones.size() == 1)
        {
            res = stones[0];
        }
        return res;
    }
};