1、题目
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.
2、思路
该题是找零问题，可以使用贪心来求解，每次找零时都先从大面值的开始找，这样用掉的钞票数量最少，能更多的满足顾客。首先对于这道题需要记录收到的钞票数量 然后在每次处理顾客找零时，都从手里已有的大面值开始找。
3、实现
public class Solution {
public boolean lemonadeChange(int[] bills) {
// 存放已收到钞票数量
int[] changeArr = new int[21];
// 钞票面值
int[] faceValue = {20, 10, 5};
for (int bill : bills){
int change = bill - 5;
changeArr[bill]++;
if (change > 0){
for (int v : faceValue){
if (change >= v && changeArr[v] > 0){
int count = change / v;
if (changeArr[v] < count){
return false;
}else{
changeArr[v]-=count;
change = change - v * count;
}
}
}
if (change != 0){
return false;
}
}
}
return true;
}
}
