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
小成202303202655 分钟前
C++~07补充
开发语言·c++
小郎君6586 分钟前
MPTT树复选框的半选状态:一棵百万节点树从递归到计数器,再到区间覆盖
算法
酌沧8 分钟前
MemT让记忆成为模型内生能力
jvm·人工智能·算法
hold?fish:palm23 分钟前
40 二叉树的直径
c++·leetcode
Shimeri1 小时前
越卷越像:我用遗传算法做了个"内卷"实验
算法·three.js·数据可视化·遗传算法·进化算法
解决小子1 小时前
【虚幻引擎】【C++】【GAS框架】构建角色基类、属性集、数值钳制
c++·笔记·游戏引擎·虚幻·gas·能力系统
Evand J1 小时前
【PMSM例程1】永磁同步电机(PMSM)转速与转子电角度观测算法|基于扩展卡尔曼滤波(EKF)的传感器状态估计与误差。附下载链接
算法·电机·1024程序员节·卡尔曼滤波
guo_wen_qiang1 小时前
二叉树的层序遍历-使用双端队列Deque
算法·leetcode·1024程序员节
爱读书的小胖2 小时前
LeetCode 55 跳跃游戏 C 语言解题分析
c语言·数据结构·算法·leetcode·1024程序员节
CCblandy2 小时前
前缀缓存命中率不靠引擎靠排列:本机仿真 3 轮,易变字段放开头 98.0%→0.0%
人工智能·python·算法·自然语言处理·性能优化