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  • 860. Lemonade Change

    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 510, or 20.
    class Solution(object):
        def lemonadeChange(self, bills):
            """
            :type bills: List[int]
            :rtype: bool
            """
            
            nums = [0] * 2
    
            for b in bills:
                if b == 5:
                    nums[0] += 1
                elif b == 10:
                    if nums[0] > 0:
                        nums[1] += 1
                        nums[0] -= 1
                    else:
                        return False
                else:
                    if nums[1] > 0 and nums[0] > 0:
                        nums[1] -= 1
                        nums[0] -= 1
                    elif nums[0] > 2:
                        nums[0] -= 3
                    else:
                        return False
    
            return True
    
            
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  • 原文地址:https://www.cnblogs.com/boluo007/p/12593691.html
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