1046_Last_Stone_Weight.cpp

/*
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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/last-stone-weight
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/

#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
		int res = 0;

		std::sort(stones.begin(), stones.end());
		while(stones.size()>=2){
			int N = stones.size();
			int x = stones[N-2], y = stones[N-1];
			if(x==y){
				stones.erase(stones.end()-1);
				stones.erase(stones.end()-1);
			}
			else{
				stones[N-2] = y-x;
				stones.erase(stones.end()-1);
				std::sort(stones.begin(), stones.end());
			}
		}
		if(stones.size()==1) res += stones[0];

		return res;
    }
};