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 weightx
is totally destroyed, and the stone of weighty
has new weighty-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
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];
}
}