[LeetCode] 860. Lemonade Change_Easy tag: Greedy

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.

 

零钱只有5, 10, 而20 可以不用管. ans = [0]*2, 分别存5的张数和10 的张数

1. if 5, ans[0] += 1

2. if 10, ans[0] -= 1

3. if 20, first if ans[1] > 0 then , ans[0] -= 1, ans[1] -= 1

　　else: ans[0] -= 3

然后看ans[0] <0, return False

else: return True

 

Code    T: O(n)   S; O(1)

class Solution(object):
    def lemonadeChange(self, bills):
        ans = [0,0]  # present 5, 10
        for money in bills:
            if money == 5:
                ans[0] += 1
            elif money == 10:
                ans[0] -= 1
                ans[1] += 1
            else:
                if ans[1]:
                    ans[1] -= 1
                    ans[0] -= 1
                else:
                    ans[0] -= 3
            if ans[0] <0 : return False
        return True