LeetCode860. Lemonade Change

文章目录

一、题目

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 do not have any change in hand at first.

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

Example 1:

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.

Example 2:

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 the change of 15 back because we only have two 10 bills.

Since not every customer received the correct change, the answer is false.

Constraints:

1 <= bills.length <= 105

billsi is either 5, 10, or 20.

二、题解

cpp 复制代码
class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int n = bills.size();
        vector<int> balance(3,0);
        for(int i = 0;i < n;i++){
            int sub = bills[i] - 5;
            if(sub == 5 && balance[0] == 0) return false;
            else if(sub == 15 && (balance[0] == 0 || (balance[1] == 0 && balance[0] < 3))) return false;
            if(sub == 5) balance[0]--;
            else if(sub == 15){
                if(balance[1] > 0) balance[0]--,balance[1]--;
                else balance[0] -= 3;
            }
            if(bills[i] == 5) balance[0]++;
            else if(bills[i] == 10) balance[1]++;
            else balance[2]++;
        }
        return true;
    }
};

更好的解法

cpp 复制代码
class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int n = bills.size();
        int five = 0,ten = 0,twenty = 0;
        for(int i = 0;i < n;i++){
            if(bills[i] == 5) five++;
            else if(bills[i] == 10){
                if(five == 0) return false;
                ten++;
                five--;
            }
            else if(bills[i] == 20){
                if(five > 0 && ten > 0) five--,ten--;
                else if(five >= 3 && ten == 0) five -= 3;
                else return false;
            }
        }
        return true;
    }
};
相关推荐
小徐不徐说1 分钟前
C++ 机器视觉入门:从零搭建 AOI 开发环境(OpenCV + CMake)
开发语言·c++·opencv
Young Doro4 分钟前
从平衡表征到最小充分因果表征:端到端神经网络因果效应估计
算法
0+11110 分钟前
算法 --递归
算法·leetcode
打工仔折腾 AI11 分钟前
介入式脑机接口的跨尺度解码与闭环调控:从算法落地到后端工程实践
人工智能·后端·python·深度学习·算法·ai agent 实战
誰能久伴不乏17 分钟前
【无标题】
c++·qt·机器人
大陈子编程31 分钟前
C++ 背包九讲:从入门到精通的动态规划专题
开发语言·c++·动态规划
凤山老林1 小时前
授权模型只会RBAC?深入 ABAC 授权模型,实现权限判断做成策略引擎
大数据·数据库·算法
starzy19901 小时前
KafkaProducer 源码深度剖析:支撑高吞吐的数据结构设计
大数据·数据结构·kafka
海绵宝宝转agent1 小时前
2026-10-21学习笔记
笔记·学习·算法
Dxy12393102161 小时前
ShardingSphere 自定义分片算法:BINARY(16) UUIDv7 哈希分片实现
算法·哈希算法