1046. Last Stone Weight

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# 題目敘述

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y . The result of this smash is:

• If x == y , both stones are destroyed, and
• If x != y , the stone of weight x is destroyed, and the stone of weight y has new weight y - x .

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone. If there are no stones left, return 0 .

# Example 1

Input: stones = [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 the last stone.

# Example 2

Input: stones = [1]
Output: 1

# 解題思路

# Solution

class Solution {

public:

    int lastStoneWeight(vector<int>& stones) {

        priority_queue<int> pq;

        for (int stone : stones) pq.push(stone);

        while (pq.size() != 1) {

            int x = pq.top();

            pq.pop();

            int y = pq.top();

            pq.pop();

            pq.push(max(x, y) - min(x, y));

        }

        return pq.top();

    }

};

import java.util.Comparator;

import java.util.PriorityQueue;

class Solution {

    public int lastStoneWeight(int[] stones) {

        if(stones.length == 0) return 0;

        PriorityQueue<Integer> pQ = new PriorityQueue<>(Comparator.reverseOrder());

        for(int s : stones){

            pQ.add(s);

        }

        while(pQ.size() >= 2){

            int first = pQ.poll();

            int second = pQ.poll();

            pQ.add(first - second);

        }

        return pQ.peek();

    }

}

class Solution:

    def lastStoneWeight(self, stones: List[int]) -> int:

        max_heap = [-stone for stone in stones]

        heapq.heapify(max_heap)

        while len(max_heap) > 1:

            x = heapq.heappop(max_heap)

            y = heapq.heappop(max_heap)

            if x != y: heapq.heappush(max_heap, x - y)

        if max_heap: return -max_heap[0]

        else: return 0