[LeetCode] 860. Lemonade Change 买柠檬找零（转）

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.

这道题说是有很多柠檬，每个卖5刀，顾客可能会提供5刀，10刀，20刀的钞票，我们刚开始的时候并没有零钱，只有收到顾客的5刀，或者 10 刀可以用来给顾客找钱，当然如果第一个顾客就给 10 刀或者 20 刀，那么是无法找零的，这里就问最终是否能够都成功找零。10 刀的钞票需要5刀的找零，20 刀的钞票可以用1张 10 刀和1张5刀，或者3张5刀的钞票。实际上我们并不需要一直保留所有钞票的个数，当某些钞票被当作零钱给了，就没有必要继续留着它们的个数了。其实上我们只关心当前还剩余的5刀和 10 刀钞票的个数，用两个变量 five 和 ten 来记录。然后遍历所有的钞票，假如遇到5刀钞票，则 five 自增1，若遇到 10 刀钞票，则需要找零5刀，则 five 自减1，ten 自增1。否则遇到的就是 20 刀的了，若还有 10 刀的钞票话，就先用 10 刀找零，则 ten 自减1，再用一张5刀找零，five 自减1。若没有 10 刀了，则用三张5刀找零，five 自减3。找零了后检测若此时5刀钞票个数为负数了，则直接返回 false，参见代码如下：

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int five = 0, ten = 0;
        for (int bill : bills) {
            if (bill == 5) ++five;
            else if (bill == 10) { --five; ++ten; }
            else if (ten > 0) { --ten; --five; }
            else five -= 3;
            if (five < 0) return false;
        }
        return true;
    }
};