1046. Last Stone Weight

• https://leetcode.com/problems/last-stone-weight/description/

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

Constraints:

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

代码

class Solution {
    public int lastStoneWeight(int[] stones) {
        int length = stones.length;
        Arrays.sort(stones);
        while(length > 1){
            int s1 = stones[stones.length - 1];
            int s2 = stones[stones.length - 2];
            if(s1 == s2){
                stones[stones.length - 1] = -1;
                stones[stones.length - 2] = -1;
                length -= 2;
            }else{
                stones[stones.length - 1] = -1;
                stones[stones.length - 2] = s1 - s2;
                length -= 1;
            }
            Arrays.sort(stones);
        }
        if(stones[stones.length - 1] != -1){
            return stones[stones.length - 1];
        }else{
            return 0;
        }
    }
}

答案

Explanation

Put all elements into a priority queue.
Pop out the two biggest, push back the difference,
until there are no more two elements left.

Complexity

Time O(NlogN)
Space O(N)

, PriorityQueue

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