LeetCode: 860. Lemonade Change

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: 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.

解题思路 —— 贪心思想

优先选择找大面额的钞票，如果遇到找不开的情况，则认为其无法找开。

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;
    }
};