LeetCode-Lemonade Change

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.

题意：有一个卖柠檬的小摊，和一个顾客队列，每一杯柠檬水的价格为5$，顾客每次会支付5$,10$或者20$；假设摊主一开始余额为0，计算是否可以对每个顾客找零；

解法：关键的地方就是要统计此时摊主手中剩余的5$,10$的数量，因为这个决定了是否足够给顾客找零；有以下几种情况：

1. 顾客支付5$，无需找零
2. 顾客支付10$，找给顾客5$
3. 顾客支付20$，找给顾客10$ + 5$，或者3张5$

对于第三种情况，在可以先支付10$的情况下应当先支付，因为5$可以使用在所有情况之中，应当保证其数量足够；

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;
    }
}