[Leetcode 1046] Last Stone Weight

原题说明

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. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

解题思路

题目要求每次取出当前石头堆中最大的两个，若它们相同，则直接销毁，若不同，则将一块重量等于它们差值的石头放入堆中。如此反复，直到最后至多只有一块石头。返回这块石头的重量，如果没有石头，返回0。

既然每次都要取出最大的两个数，我们自然想到利用最大堆来维护数据。现将所有石头放入最大堆中，然后每次取出堆顶两个数进行操作。直到堆的规模小于等于1。如果等于1，返回堆顶元素，否则返回0。

示例代码 (cpp)

class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue<int> pq;
        for (auto& stone : stones) {
            pq.push(stone);
        }
        while (pq.size() >= 2) {
            int t1 = pq.top();
            pq.pop();
            int t2 = pq.top();
            pq.pop();
            if (t1 != t2) {
                pq.push(t1 - t2);
            }
        }
        if (pq.empty()) {
            return 0;
        }
        return pq.top();
    }
};

示例代码 (java)

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b)-> b - a);
        for (int a : stones)
            pq.offer(a);
        while (pq.size() > 1)
            pq.offer(pq.poll() - pq.poll());
        return pq.poll();
    }
}

示例代码 (python)

class Solution:
    def lastStoneWeight(self, stones):
        h = [-x for x in stones]
        heapq.heapify(h)
        while len(h) > 1 and h[0] != 0:
            heapq.heappush(h, heapq.heappop(h) - heapq.heappop(h))
        return -h[0]

复杂度分析

时间复杂度: 假设一共有N数，建立堆后，每次取出两个数，加入一个数，直到堆得规模小于等于1。一共有2N次删除和N次加入。每次删除时间复杂度是O(logN), 每次加入是O(1)。所有总的时间复杂度是O(NlogN)
空间复杂度: O(N)

归纳总结

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