QUESTION:

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

EXPLANATION:

其实我本来也没有想到这道题会这么简单。以为起码会有一个时间复杂度或者空间复杂度的限制。但是并没有。那么就比较简单了。

1.将数组排序

2.将最大的两个数进行smash

3.结果保存在倒数第二个

4.截取数组

5.反复进行到只剩最后一位。

可以优化的点：如果x==y，那么其实可以直接将两者都去掉。

SOLUTION:

class Solution {
    public int lastStoneWeight(int[] stones) {
        while (stones.length!=1){
            Arrays.sort(stones);
            stones[stones.length-2] = stones[stones.length-1]-stones[stones.length-2];
            stones = Arrays.copyOf(stones,stones.length-1);
        }
        return stones[0];
    }
}