Description： 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.

題意：有一個賣檸檬的小攤，和一個顧客佇列，每一杯檸檬水的價格為5$，顧客每次會支付5$,10$或者20$；假設攤主一開始餘額為0，計算是否可以對每個顧客找零；

解法：關鍵的地方就是要統計此時攤主手中剩餘的5$,10$的數量，因為這個決定了是否足夠給顧客找零；有以下幾種情況：

1. 顧客支付5$，無需找零
2. 顧客支付10$，找給顧客5$
3. 顧客支付20$，找給顧客10$ + 5$，或者3張5$

對於第三種情況，在可以先支付10$的情況下應當先支付，因為5$可以使用在所有情況之中，應當保證其數量足夠；

Java

class Solution {
    public boolean lemonadeChange(int[] bills) {
        int fiveCnt = bills[0] == 5 ? 1 : 0;
        int tenCnt = bills[0] == 10 ? 1 : 0;
        for (int i = 1; i < bills.length; i++) {
            if (bills[i] == 5) {
                fiveCnt++;
            } else if (bills[i] == 10 && fiveCnt > 0) {
                fiveCnt--;
                tenCnt++;
            } else if (fiveCnt > 0 && tenCnt > 0){
                fiveCnt--;
                tenCnt--;
            } else if (fiveCnt > 2) {
                fiveCnt -= 3;
            } else {
                return false;
            }
        }
        return true;
    }
}