我们看一道比较简单的题，作为今天刷题的开胃菜。

柠檬水找零

题目

在柠檬水摊上，每一杯柠檬水的售价为 5 美元。顾客排队购买你的产品，（按账单 bills 支付的顺序）一次购买一杯。

每位顾客只买一杯柠檬水，然后向你付 5 美元、10 美元或 20 美元。你必须给每个顾客正确找零，也就是说净交易是每位顾客向你支付 5 美元。

注意，一开始你手头没有任何零钱。

给你一个整数数组 bills ，其中 bills[i] 是第 i 位顾客付的账。如果你能给每位顾客正确找零，返回 true ，否则返回 false 。

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.

Given an integer array bills where bills[i] is the bill the ith customer pays, return true if you can provide every customer with correct change, or false otherwise.

Example:

Input: bills = [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.

Input: bills = [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 not give change of $15 back because we only have two $10 bills.
Since not every customer received correct change, the answer is false.

Input: bills = [5,5,10]
Output: true

Input: bills = [10,10]
Output: false

Constraints:

1 <= bills.length <= 105
bills[i] is either 5, 10, or 20.

思考线

解题思路

根据题意，我们可以记录下5美元和10美元的数量。

在找零的时候我们能找10美元就不用找5美元，直到结束或者发现不能找零为止。

根据上面的理解我们可以得到代码如下

/**
 * @param {number[]} bills
 * @return {boolean}
 */
var lemonadeChange = function (bills) {
    // 记录 5, 10美元的数量，能找10美元不用5美元，直到结束，或者发现不能找零。
    const hash = {
        five: 0,
        ten: 0,
    }
    for (let i = 0; i < bills.length; i++) {
        const item = bills[i];
        if (item === 5) {
            hash.five++
        } else if (item === 10) {
            hash.ten++;
            hash.five--;
            if (hash.five < 0) return false;
        } else {
            if (hash.ten > 0) {
                hash.ten--;
                hash.five--;
            } else {
                hash.five -=3;
            }
            if (hash.five < 0 || hash.ten < 0) return false;
        }
    }
    return true;
};

复杂度分析

时间复杂度：O(N)，其中 N 是 bills 的长度。
空间复杂度：O(1)。

[路飞]_每天刷leetcode_15(柠檬水找零 Lemonade Change)

柠檬水找零

题目

解题思路