Coding

1046. Last Stone Weight

2023-04-24 2 min read
#Array#Heap (Priority Queue)
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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

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();
    }
}
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