【Rikko】860. Change for Lemonade
At the lemonade stand, a glass of lemonade sells for $5. Customers line up to buy your products, one cup at a time (in the order bills are paid).
Each customer buys just one glass of lemonade and pays you $5, $10, or $20. You have to give each customer the correct change, which means the net transaction is that each customer pays you $5.
Note that you don't have any change on hand at first.
You are given an integer array bills, where bills[i] is the bill paid by the i-th customer. Return true if you were able to give each customer correct change, otherwise return false.
Example 1:
Input : bills = [5,5,5,10,20]
Output : true
Explanation :
For the first 3 customers, we collect 3 $5 bills in sequence. For the 4th customer, we take a $10 bill and give back $5. With the 5th customer, we return a $10 bill and a $5 bill. Since all customers got correct change, we output true.
Example 2:
Input : bills = [5,5,10,10,20]
Output : false
Explanation :
For the first 2 customers, we collect 2 $5 bills in sequence. For the next 2 customers, we take a $10 bill and give back $5. For the last customer, we can't refund the $15 because we only have two $10 bills right now. Since not every customer gets the correct change, the answer is false.
Tips :
1 <= bills.length <= 1 0 5 10^5 105
bills[i] is either 5 or 10 or 20
answer
Greedy thinking, in the process of finding change, first look for the one with a larger denomination, and then find the one with a smaller denomination
class Solution {
public boolean lemonadeChange(int[] bills) {
int five = 0, ten = 0;
for (int bill : bills) {
//五块
if (bill == 5) {
five++;//直接收下
}
//十块
else if (bill == 10) {
if (five == 0) {
return false;//无法找零
}
five--;
ten++;
}
//二十
else {
//当可以正确找零时，两种找零的方式中我们更倾向于第一种，因为需要使用 5美元的找零场景会比需要使用 10美元的找零场景多
if (five > 0 && ten > 0) {
five--;
ten--;
}
//找5*3
else if (five >= 3) {
five -= 3;
}
else {
return false;
}
}
}
return true;
}
}