## 题目描述

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: trueExplanation: 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: falseExplanation: 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 <= 10000bills[i] will be either 5, 10, or 20.`

## AC 代码

`class Solution {public:    bool lemonadeChange(vector<int>& bills) {        int cnt[3] = {0};  // 0, 1, 2 ==> 5, 15, 20        for(int i = 0; i < bills.size(); ++i)        {            if(bills[i] == 5)            {                ++cnt[0];            }            else if(bills[i] == 10) // 找 5 块            {                ++cnt[1];                --cnt[0];            }            else if(bills[i] == 20) // 找 15 块            {                ++cnt[2];                if(cnt[1] > 0) // 优先找 10 + 5                {                    --cnt[1];                     --cnt[0];                }                else // 没有 10 块，直接找5块                {                    cnt[0] -= 3;                }            }            if(cnt[0] < 0) // 剩余 5 块钱面额钞票数量为负，则找不开            {                return false;            }        }        return true;    }};`