LeetCode 2034. 股票价格波动

2034. 股票价格波动

Solution

思路一:有序集合+哈希表

查询最新股票价格:维护最大的时间戳 哈希表直接查询。

最高,最低价格查询:维护股票价格的有序集合。

由于存在不同时间戳有相同的股票价格,因此股票价格需要记录出现次数,更新的时候维护即可。

class StockPrice {
    int curTimestamp;
    HashMap<Integer, Integer> timePricestamp;
    TreeMap<Integer, Integer> priceCnt;
    public StockPrice() {
        curTimestamp = 0;
        timePricestamp = new HashMap<>();
        priceCnt = new TreeMap<>();
    }
    
    public void update(int timestamp, int price) {
        curTimestamp = Math.max(curTimestamp, timestamp);
        if (timePricestamp.containsKey(timestamp)) {
            int oldPrice = timePricestamp.get(timestamp);
            int cnt = priceCnt.get(oldPrice);
            if (cnt == 1) {
                priceCnt.remove(oldPrice);
            } else {
                priceCnt.put(oldPrice, cnt - 1);
            }
        } 
        priceCnt.put(price,  priceCnt.getOrDefault(price, 0) + 1);
        timePricestamp.put(timestamp, price);
        
    }
    
    public int current() {
        return timePricestamp.get(curTimestamp);
    }
    
    public int maximum() {
        return priceCnt.lastKey();
    }
    
    public int minimum() {
        return priceCnt.firstKey();
    }
}

/**
 * Your StockPrice object will be instantiated and called as such:
 * StockPrice obj = new StockPrice();
 * obj.update(timestamp,price);
 * int param_2 = obj.current();
 * int param_3 = obj.maximum();
 * int param_4 = obj.minimum();
 */

思路二:哈希表+两个优先级队列

如果更新的时候直接取最大值,最小值就会出现依然保存历史失效的股票价格,因此在更新的时候我们可以把所有更新数据存入最大堆,最小堆,保存时间与股票价格的对应关系,在查询最大、最小的时候,我们就从最大最小堆里查找,同时维护。如果出现和哈希表中的数据一致时,说明数据正常,不一致就是失效数据。

class StockPrice {
    int curTimestamp;
    HashMap<Integer, Integer> stocks;
    PriorityQueue<int[]> maxPriority;
    PriorityQueue<int[]> minPriority;
    public StockPrice() {
        curTimestamp = 0;
        stocks = new HashMap<>();
        maxPriority = new PriorityQueue<int[]>((a, b) -> b[0] - a[0]);
        minPriority = new PriorityQueue<int[]>((a, b) -> a[0] - b[0]);
    }
    
    public void update(int timestamp, int price) {
        curTimestamp = Math.max(timestamp, curTimestamp);
        stocks.put(timestamp, price);
        maxPriority.offer(new int[]{price, timestamp});
        minPriority.offer(new int[]{price, timestamp});
    }
    
    public int current() {
        return stocks.get(curTimestamp);
    }
    
    public int maximum() {
        while (true) {
            int[] vals = maxPriority.peek();
            int price = vals[0], timestamp = vals[1];
            if (stocks.get(timestamp) == price) {
                return price;
            }
            maxPriority.poll();
        }
    }

    public int minimum() {
        while (true) {
            int[] vals = minPriority.peek();
            int price = vals[0], timestamp = vals[1];
            if (stocks.get(timestamp) == price) {
                return price;
            }
            minPriority.poll();
        }
    }
}

/**
 * Your StockPrice object will be instantiated and called as such:
 * StockPrice obj = new StockPrice();
 * obj.update(timestamp,price);
 * int param_2 = obj.current();
 * int param_3 = obj.maximum();
 * int param_4 = obj.minimum();
 */
posted @ 2022-01-23 15:49  Frontierone  阅读(64)  评论(0编辑  收藏  举报