【LeetCode】860. Lemonade Change 解题报告(Python)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/lemonade-change/description/
题目描述
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
日期
2018 年 7 月 4 日 ———— 夏天挺热的,记得吃饭,防止低血糖
2018 年 11 月 13 日 —— 时间有点快
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