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.

方法1：暴力求解（5元是关键，能省则省；10元能用就用）

遍历数组，存储5元10元个数。

1.遇到5元，5元个数+1

2.遇到10元，10元个数+1,5元个数-1

3.遇到20元，如果有10元，10元个数-1,5元个数-1，如果没有10元，5元个数-3

判断，如果5元个数小于0，则返回false。遍历结束5元个数>=,返回true.

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int billKind[2]={0};
        for(int a :bills){
            if(a==5) billKind[0]++;
            if(a==10){
                billKind[1]++;
                billKind[0]--;
            }
            if(a==20){
                if(billKind[1]>0){
                    billKind[1]--;
                    billKind[0]--;
                }else{
                    billKind[0]-=3;
                }  
            }
            if(billKind[0]<0) return false;
        }
        return true;
    }
};