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.
这是一个关于找零的问题。首先，因为初始没有资本，而且每份柠檬水的售价是5美元。因此，5美元是最“珍贵”的找零资源。
如果顾客给的是10美元，那么只能找给他一张5美元的。
如果顾客给的是20美元，因为5美元比较稀缺，因此如果有10美元的钞票，则优先找给他10美元，然后再找一张5美元；否则的话，就直接给三张5美元的钞票。
如果5美元的数量变为“负”，说明找零失败，返回False；否则就返回True。
class Solution {
public boolean lemonadeChange(int[] bills) {
Map<Integer, Integer> map = new HashMap<>();
map.put(5,0);
map.put(10,0);
for(int bill : bills){
if(bill == 5){
map.put(5, map.get(5)+1);
}
else if(bill == 10){
if(map.get(5) == 0){
return false;
}
map.put(5,map.get(5)-1);
map.put(10,map.get(10)+1);
}
else if(bill == 20){
if(map.get(5) == 0){
return false;
}
if(map.get(10) == 0 && map.get(5) < 3){
return false;
}
if(map.get(10) > 0 && map.get(5) == 0){
return false;
}
if(map.get(10) > 0){
map.put(10, map.get(10)-1);
map.put(5, map.get(5)-1);
}
else{
map.put(5,map.get(5)-3);
}
}
}
return true;
}
}