堆&&优先队列&&TreeMap

题目描述

5710. 积压订单中的订单总数

题解

题目不难，主要是要读懂题意，一步步模拟，代码较长，需要细心检查。
坑较多，比如我犯了很多傻逼问题：想都不想就拿1<<9+7当作1000000007,更傻逼的是,<<的优先级低于+号,<<都没用对。
实时取最大和最小，可以用堆或者优先队列实现。这里我使用了Java的TreeMap。

class Solution {
    public static int getNumberOfBacklogOrders(int[][] orders) {
        TreeMap<Integer,Long> sells = new TreeMap<>();//价格：数量
        TreeMap<Integer,Long> buys = new TreeMap<>();//价格：数量
        for(int i=0;i<orders.length;i++){
            if(orders[i][2]==0){//buy
                long curBuyAmount = orders[i][1];
                int buyPrice = orders[i][0];
                while (curBuyAmount>0){
                    if(sells.isEmpty()){
                        buys.put(buyPrice,buys.getOrDefault(buyPrice,0l)+curBuyAmount);//添加剩余的订单到积压订单
                        break;
                    }
                    if(sells.firstKey()<=buyPrice){
                        if(curBuyAmount<sells.get(sells.firstKey())){
                            sells.put(sells.firstKey(),sells.get(sells.firstKey())-curBuyAmount);
                            curBuyAmount = 0;
                        }else{
                            curBuyAmount -= sells.get(sells.firstKey());
                            sells.remove(sells.firstKey());
                        }
                    }else{
                        buys.put(buyPrice,buys.getOrDefault(buyPrice,0l)+curBuyAmount);//添加剩余的订单到积压订单
                        break;
                    }
                }
            }else{//sell
                long curSellAmount = orders[i][1];
                int sellPrice = orders[i][0];
                while (curSellAmount>0){
                    if(buys.isEmpty()){
                        sells.put(sellPrice,sells.getOrDefault(sellPrice,0l)+curSellAmount);
                        break;
                    }
                    if(buys.lastKey()>=sellPrice){
                        if(curSellAmount<buys.get(buys.lastKey())){
                            buys.put(buys.lastKey(),buys.get(buys.lastKey())-curSellAmount);
                            curSellAmount = 0;
                        }else{
                            curSellAmount -= buys.get(buys.lastKey());
                            buys.remove(buys.lastKey());
                        }
                    }else{
                        sells.put(sellPrice,sells.getOrDefault(sellPrice,0l)+curSellAmount);//添加剩余的订单到积压订单
                        break;
                    }
                }
            }
        }
        long mod = 1000000007;
        long overStockNum = 0;
        for(Long i:buys.values()){
            overStockNum = (overStockNum + i)%mod;//实时取模
        }
        for(Long i:sells.values()){
            overStockNum = (overStockNum + i)%mod;
        }
        return (int)(overStockNum%mod);
    }
}