1046. Last Stone Weight

#### 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`

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。

#### 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];
}
}
``````
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