1046 Last Stone Weight 最后一块石头的重量
Description:
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:
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
题目描述:
有一堆石头，每块石头的重量都是正整数。
每一回合，从中选出两块最重的石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
如果 x == y，那么两块石头都会被完全粉碎；
如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
最后，最多只会剩下一块石头。返回此石头的重量。如果没有石头剩下，就返回 0。
提示：
1 <= stones.length <= 30
1 <= stones[i] <= 1000
思路:
- 按照题目要求, 一直排序, 将最大的两个石头碰撞
时间复杂度O(n ^ 2lgn), 空间复杂度O(1) - 由于每次都是对最大的两块石头进行操作, 可以采用大顶堆
时间复杂度O(nlgn), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
int lastStoneWeight(vector& stones)
{
priority_queue pq(stones.begin(), stones.end());
while (pq.size() > 1)
{
int first = pq.top();
pq.pop();
int second = pq.top();
pq.pop();
if (first - second) pq.push(first - second);
}
return pq.empty() ? 0 : pq.top();
}
};
Java:
class Solution {
public int lastStoneWeight(int[] stones) {
int n = stones.length;
if (n == 1) return stones[0];
if (n == 2) return Math.abs(stones[0] - stones[1]);
Arrays.sort(stones);
if (stones[n - 3] == 0) return stones[n - 1] - stones[n - 2];
stones[n - 1] -= stones[n - 2];
stones[n - 2] = 0;
return lastStoneWeight(stones);
}
}
Python:
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
while len(stones) > 1:
stones.sort()
if stones[-1] == stones[-2]:
stones = stones[:-2]
else:
stones[-2] = stones[-1] - stones[-2]
stones.pop(-1)
return stones[0] if stones else 0