题目地址：https://leetcode.com/problems/lemonade-change/description/
题目描述
Ata 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.
题目大意
假如你是卖柠檬水的，现在要给顾客找零钱。一杯柠檬水的售价是5刀，顾客给的钱有5刀，10刀，20刀三种情况。刚开始的时候柜台没有零钱。每个顾客买一杯水。判断给出bills的情况下，能否完成卖柠檬水找零钱的任务。
解题方法
这个一看就是很明显的贪心算法，其实大家都知道，如果要找零钱的话，肯定先按照给大的零钱开始，不够再用小的零钱。
这个题目还是有点简单，题目中的零钱其实只有两种：5块和10块。
当给的是10块的时候，肯定找一张5块的就够了。 当给的是20块的时候，如果有10块的，先找10块的零钱，然后再找一张5块的。如果没有10块的，只能找三张5块的了。
代码如下：
class Solution:
def lemonadeChange(self, bills):
"""
:type bills: List[int]
:rtype: bool
"""
changes = {5:0, 10:0}
for bill in bills:
if bill == 5:
changes[5] += 1
elif bill == 10:
if changes[5] == 0:
return False
else:
changes[10] += 1
changes[5] -= 1
elif bill == 20:
if changes[10] != 0:
if changes[5] == 0:
return False
else:
changes[5] -= 1
changes[10] -= 1
else:
if changes[5] < 3:
return False
else:
changes[5] -= 3
return True
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