Leetcode 860 Lemonade Change

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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:

  1. 0 <= bills.length <= 10000
  2. bills[i] will be either 5, 10, or 20.

分析：

  1. 一个物品5美元，每个人依次来买，且只买一个
  2. 你初始没钱，要保证每个顾客你都能立即完成找钱操作
  3. 顾客给的钱只会是5,10,20，数据长度达到10000，说明O(n^2)复杂度不可行

思路：

  1. 如果顾客给的是5元，那么不用做任何操作，直接收下即可
  2. 如果顾客给的是10元，那么需要找一个5元
  3. 如果顾客给的是20元，那么我们有两种选择，一种是找一个5元和一个10元，一种是找三个5元，那么显然因为5元更加灵活（可以用来应对10元情况），所以我们应当优先用第一种找钱方法。
  4. 用一个字典记录5元，10元的个数，当剩余钱数不够找即return False!
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class Solution(object):
    def lemonadeChange(self, bills):
        """
        :type bills: List[int]
        :rtype: bool
        """

        d = {5:0,10:0}
        for val in bills:
            # 5元情况
            if val == 5:
                d[5] += 1
            # 10元情况，找一个5元
            elif val == 10:
                if d[5]:
                    d[5] -= 1
                    d[10] += 1
                else:
                    return False
            # 20元情况，优先用一个10元和一个5元找钱
            else:
                if d[10] and d[5]:
                    d[10] -= 1
                    d[5] -= 1
                elif d[5] >= 3:
                    d[5] -= 3
                else:
                    return False
        return True

45 / 45 test cases passed.
difficulty: easy
Runtime: 99 ms