# Leetcode 860. Lemonade Change 練習
> Difficulty : `Easy`
> `Java` `演算法Algorithm`

> 撰寫人[name=KVJK_2125] [time=SAT, May 18, 2024 22:30]

## 題目
### 說明 Description
EN：<br>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 do not have any change in hand at first.

Given an integer array `bills` where `bills[i]` is the bill the `ith` customer pays, return `true` if you can provide every customer with the correct change, or `false` otherwise.

繁中：<br>在檸檬水攤上，每一杯檸檬水的售價為`5`美元。顧客排隊購買你的產品，（按帳單`bills`支付的順序）一次購買一杯。
<br>每位顧客只買一杯檸檬水，然後向你付`5`美元、`10`美元或`20`美元。你必須給每個顧客正確找零，這意味著淨交易是每位顧客向你支付`5`美元。
<br>
注意，一開始你手邊沒有任何零錢。
<br>
給你一個整數陣列`bills`，其中`bills[i]`是第`i`位顧客付的帳，如果你能給每位顧客正確找零，回傳`truw`，否則回傳`false`。

### 範例 Example
Example 1:
> Input: bills = [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: bills = [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 not give the change of $15 back because we only have two $10 bills.
Since not every customer received the correct change, the answer is false.

### 限制條件 Constraints
- `1 <= bills.length <= 105`
- `bills[i]` is either `5`, `10`, or `20`.

## 解題
### 思路 Intuition/Approach

- Step1.<br> 建立變數`coin5`和`coin10`紀錄手中擁有的零錢，初始設定為0

- Step2.<br> 建立for迴圈，設定為`bills`長度的次數

- Step3.<br> 若收到5元，則`coin5++`；<br><br>若收到10元，則確認是否能夠找零：<br>- 是：`coin5--`且`coin10++`<br>- 否：不可找零，傳回`false`<br><br>若收到20元，則確認是否能夠找零：<br>- 有5元及10元：`coin5--`且`coin10--`<br>- 只有5元，且5元最少有三個：`coin5-=3`<br>- 以上情況皆非：不可找零，傳回false<br><br> 若全部順利找零，則最終回傳`true`。
### 程式碼 Code（加註解）

class Solution {
    public boolean lemonadeChange(int[] bills) {
        //建立變數紀錄手上有的零錢
        int coin5 = 0,coin10 = 0;
        //開始交易（？）
        for(int i : bills)
        {
            //如果是5，則5元+1
            if(i == 5)
                coin5 ++;
            //如果是10，則確認有無零錢    
            else if(i == 10)
            {
                //可找錢的情況
                if(coin5 > 0)
                {
                    coin5--;
                    coin10++;
                }
                //不可找錢
                else
                    return false;
            }
            //因為只有5、10、20，所以這邊是20
            else
            {
                //可找錢的情況：5元跟10元都有，則找錢
                if(coin5 > 0 && coin10 > 0)
                {
                    coin5 --;
                    coin10 --;
                }
                //可找錢的情況：沒有10元，5元有兩個以上
                else if(coin5 > 2)
                {
                    coin5 -= 3;
                }
                //不可找錢的情況
                else
                    return false;
            }
        }
        return true;
    }
}