Media
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https://leetcode.com/problems...

class StockPrice {    // key: time, value: price    private Map<Integer, Integer> timePrice;    // key: price, value: frequency,应用tree map,使得依照price从小到大排序,不便间接获取max / min    private TreeMap<Integer, Integer> priceFreq;    private int latestTime; // 最新的工夫    public StockPrice() {        timePrice = new HashMap<>();        latestTime = 0;        priceFreq = new TreeMap<>();    }    public void update(int timestamp, int price) {        // 更新最新工夫,始终让latestTime放弃为最近的工夫戳,在调用current办法时,间接从map中获取        latestTime = Math.max(latestTime, timestamp);        // 如果工夫戳之前保留过,那么对频率表进行更新        if (timePrice.containsKey(timestamp)) {            // 依据工夫戳,拿到之前的旧price            int oldPrice = timePrice.get(timestamp);            // 因为更改了,所以把频率-1            priceFreq.put(oldPrice, priceFreq.get(oldPrice) - 1);            // 如果该price频率更改后为0,那么须要移除该price            if (priceFreq.get(oldPrice) == 0) priceFreq.remove(oldPrice);        }        // 别离更新两个map中的新的price        timePrice.put(timestamp, price);        priceFreq.put(price, priceFreq.getOrDefault(price, 0) + 1);    }    public int current() {        // 依据最新工夫间接获取price        return timePrice.get(latestTime);    }    public int maximum() {        // 从priceFreq中间接取最初那个最大price        return priceFreq.lastKey();    }    public int minimum() {        // 从priceFreq中间接取第一个最小的price        return priceFreq.firstKey();    }}