1046. Last Stone Weight - Easy

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 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

 

模拟过程即可，time = O(nlogn), space = O(n)

naive solution:

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        for(int stone : stones) {
            maxHeap.offer(stone);
        }
        while(maxHeap.size() >= 2) {
            int x = maxHeap.poll();
            int y = maxHeap.poll();
            if(x != y) {
                maxHeap.offer(Math.abs(y - x));
            }
        }
        return maxHeap.size() > 0 ? maxHeap.peek() : 0;
    }
}

 

优化代码：

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        for(int stone : stones) {
            maxHeap.offer(stone);
        }
        while(maxHeap.size() >= 2) {
            maxHeap.offer(maxHeap.poll() - maxHeap.poll());
        }
        return maxHeap.peek();
    }
}