[LeetCode] Lemonade Change

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题目描述

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.

Example 2:

Input: [5,5,10]
Output: true

Example 3:

Input: [10,10]
Output: false

Example 4:

Input: [5,5,10,10,20]
Output: false
Explanation:
From the first two customers in order, we collect two \$5 bills.
For the next two customers in order, we collect a \$10 bill and give back a \$5 bill.
For the last customer, we can’t give change of \$15 back because we only have two \$10 bills.
Since not every customer received correct change, the answer is false.

Note:

0 <= bills.length <= 10000
bills[i] will be either 5, 10, or 20.

有一个柠檬水摊，每瓶柠檬水售价5刀，货主只接收 5，10，20 的现金，问题是摊主是否可以按照购买的顺序完成找零。
解决思路是使用两个变量分别记录5元现金和10元现金的张数，每次交易的时候判断是否可以完成找零并更新现金数目

C++实现

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class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        unsigned five_cash_cnt = 0;
        unsigned ten_cash_cnt = 0;

        for (const int bill : bills)
        {
            switch (bill)
            {
                case 5:
                    five_cash_cnt++;
                    break;
                case 10:
                    if (!five_cash_cnt)
                    {
                        return false;
                    }
                    five_cash_cnt--;
                    ten_cash_cnt++;
                    break;
                case 20:
                    if (ten_cash_cnt && five_cash_cnt)
                    {
                        ten_cash_cnt--;
                        five_cash_cnt--;
                    }
                    else if (five_cash_cnt > 2)
                    {
                        five_cash_cnt -= 3;
                    }
                    else
                    {
                        return false;
                    }
                    break;
            }
        }
        return true;
    }
};
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