leetcode 860. Lemonade Change

最新推荐文章于 2025-07-23 22:59:22 发布

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最新推荐文章于 2025-07-23 22:59:22 发布

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分类专栏： leetcode 文章标签：算法

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本文探讨了一个经典的找零问题——柠檬水摊位的找零挑战。顾客排队购买每杯5美元的柠檬水，使用5美元、10美元或20美元的钞票支付。文章详细解释了如何确保每位顾客都能收到正确的找零，并提供了一种有效的算法实现。

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At a lemonade stand, each lemonade costs $5.

Customers are standing in a queue to buy from you, and order one at a time (in the order specified by bills).

Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill. You must provide the correct change to each customer, so that the net transaction is that the customer pays $5.

Note that you don’t have any change in hand at first.

Return true if and only if you can provide every customer with correct change.

Example 1:

Input: [5,5,5,10,20]
Output: true
Explanation:
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.

找零钱的问题，柠檬水一杯5块，刚开始手上没有钱，按顺序从数组得到零钱，问是否能给数组中每个数找零

思路：
比较简单的是只有3中面值的货币，5，10和20，显然5是不用找零的，10的话只需要找5，20的话要么找一张10一张5，要么3张5
找零方法是固定的，所以只需要保存3中面值货币的数量，一旦需要的时候数量<=0, 直接返回false

//2ms
    public boolean lemonadeChange(int[] bills) {
        if (bills == null || bills.length == 0) {
            return true;
        }
        
        int[] billCounts = new int[3]; //5, 10, 20
        
        for (int i = 0; i < bills.length; i++) {
            switch (bills[i]) {
                case 5:
                    billCounts[0] ++;
                    break;
                case 10:
                    if (billCounts[0] <= 0) {
                        return false;
                    }
                    billCounts[0] --;
                    billCounts[1] ++;
                    break;
                case 20:
                    if ((billCounts[1] <= 0 && billCounts[0] < 3) ||
                        (billCounts[0] <= 0)){
                        return false;
                    }
                    
                    if (billCounts[1] <= 0) {
                        billCounts[0] -= 3;
                    } else {
                        billCounts[1] --;
                        billCounts[0] --;
                    }
                    billCounts[2] ++;
                    break;
            }
        }
        
        return true;
    }