【LeetCode】682. Baseball Game

本文介绍了一种基于数组实现的棒球游戏计分算法,使用堆栈来记录每轮比赛得分并支持特殊操作,包括取消上一轮得分、加倍得分及累加最近两轮得分等功能。

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Python版本:python3.6.2


682. Baseball Game


You're now a baseball game point recorder.
Given a list of strings, each string can be one of the 4 following types:
Integer (one round's score): Directly represents the number of points you get in this round.
"+" (one round's score): Represents that the points you get in this round are the sum of the last two valid round's points.
"D" (one round's score): Represents that the points you get in this round are the doubled data of the last valid round's points.
"C" (an operation, which isn't a round's score): Represents the last valid round's points you get were invalid and should be removed.
Each round's operation is permanent and could have an impact on the round before and the round after.
You need to return the sum of the points you could get in all the rounds.


Example 1:
Input: ["5","2","C","D","+"]
Output: 30
Explanation:
Round 1: You could get 5 points. The sum is: 5.
Round 2: You could get 2 points. The sum is: 7.
Operation 1: The round 2's data was invalid. The sum is: 5.
Round 3: You could get 10 points (the round 2's data has been removed). The sum is: 15.
Round 4: You could get 5 + 10 = 15 points. The sum is: 30.


Example 2:
Input: ["5","-2","4","C","D","9","+","+"]
Output: 27
Explanation:
Round 1: You could get 5 points. The sum is: 5.
Round 2: You could get -2 points. The sum is: 3.
Round 3: You could get 4 points. The sum is: 7.
Operation 1: The round 3's data is invalid. The sum is: 3.
Round 4: You could get -4 points (the round 3's data has been removed). The sum is: -1.
Round 5: You could get 9 points. The sum is: 8.
Round 6: You could get -4 + 9 = 5 points. The sum is 13.
Round 7: You could get 9 + 5 = 14 points. The sum is 27.


Note:
The size of the input list will be between 1 and 1000.
Every integer represented in the list will be between -30000 and 30000.

先贴自己的代码:

class Stack:
    def __init__(self):
        self.items = []

    def push(self, item):
        self.items.append(item)

    def pop(self):
        return self.items.pop()

    def size(self):
        return len(self.items)

    def top(self):
        return self.items[self.size() - 1]

class Solution:
    def calPoints(self, ops):
        """
        :type ops: List[str]
        :rtype: int
        """
        a = Stack()
        total = 0
        for operation in ops:
            if (operation[0] == '-' and operation[1:].isdigit()) or operation.isdigit():
                a.push(int(operation))
            if operation == "C":
                a.pop()
            if operation == "D":
                a.push(a.top() * 2)
            if operation == "+":
                a.push(a.items[-1]+a.items[-2])
        for i in a.items:
            total += i
        return total
验证下:

x = Solution()
print(x.calPoints(ops=["5","2","C","D","+"]))


这次用到了堆栈,记得上次二叉树的题目也是用堆栈,我硬是没搞懂二叉树,所有就放弃了,这次再次遇到类似需要用到堆栈的题目,想明白了,堆栈这类题目都是基于数组的操作,堆栈可以自己用python定义:

class Stack:
    def __init__(self):
        self.items = []

    def push(self,item):
        self.items.append(item)

    def pop(self):
        return self.items.pop()

    def clear(self):
        del self.items[:]

    def empty(self):
        return self.size() == 0

    def size(self):
        return len(self.items)

    def top(self):
        return self.items[self.size()-1]
其中,

Stack():创建堆栈

push(item):向栈顶插入项

pop():返回栈顶的项,并从堆栈中删除该项

clear():清空堆栈

empty():判断堆栈是否为空

size():返回堆栈中项的个数

top():返回栈顶的项

--------------------------------------------以下大神解法--------------------------------------------------

class Solution(object):
    def calPoints(self, ops):
        # Time: O(n)
        # Space: O(n)
        history = []
        for op in ops:
            if op == 'C':
                history.pop()
            elif op == 'D':
                history.append(history[-1] * 2)
            elif op == '+':
                history.append(history[-1] + history[-2])
            else:
                history.append(int(op))
        return sum(history)

思想是一样的,差别在于速度,elseif可以加快算法速度,sum()函数取代for循环也可以加速算法。同时也可以看出,堆栈跟数组之间的区别,堆栈是在数组的基础上,多了FILO的性质,pop()函数。


  • 函数sum():可用于可迭代对象的相加,如数组。

### 如何在 VSCode 中安装和配置 LeetCode 插件以及 Node.js 运行环境 #### 安装 LeetCode 插件 在 VSCode 的扩展市场中搜索 `leetcode`,找到官方提供的插件并点击 **Install** 按钮进行安装[^1]。如果已经安装过该插件,则无需重复操作。 #### 下载与安装 Node.js 由于 LeetCode 插件依赖于 Node.js 环境,因此需要下载并安装 Node.js。访问官方网站 https://nodejs.org/en/ 并选择适合当前系统的版本(推荐使用 LTS 版本)。按照向导完成安装流程后,需确认 Node.js 是否成功安装到系统环境中[^2]。 可以通过命令行运行以下代码来验证: ```bash node -v npm -v ``` 上述命令应返回对应的 Node.js 和 npm 的版本号。如果没有正常返回版本信息,则可能未正确配置环境变量。 #### 解决环境路径问题 即使完成了 Node.js 的安装,仍可能出现类似 “LeetCode extension needs Node.js installed in environment path” 或者 “command ‘leetcode.toggleLeetCodeCn’ not found” 的错误提示[^3]。这通常是因为 VSCode 未能识别全局的 Node.js 路径或者本地安装的 nvm 默认版本未被正确加载[^4]。 解决方法如下: 1. 手动指定 Node.js 可执行文件的位置 在 VSCode 设置界面中输入关键词 `leetcode`,定位至选项 **Node Path**,将其值设为实际的 Node.js 安装目录下的 `node.exe` 文件位置。例如:`C:\Program Files\nodejs\node.exe`。 2. 使用 NVM 用户管理工具调整默认版本 如果通过 nvm 工具切换了不同的 Node.js 版本,请确保设置了默认使用的版本号。可通过以下指令实现: ```bash nvm alias default <version> ``` 重新启动 VSCode 后测试功能键是否恢复正常工作状态。 --- #### 配置常用刷题语言 最后一步是在 VSCode 设置面板中的 LeetCode 插件部分定义个人习惯采用的主要编程语言作为默认提交方式之一。这样可以减少频繁修改编码风格的时间成本。 --- ### 总结 综上所述,要在 VSCode 上顺利启用 LeetCode 插件及其关联服务,除了基本插件本身外还需额外准备支持性的后台框架——即 Node.js 应用程序引擎;同时针对特定场景下产生的兼容性障碍采取针对性措施加以修正即可达成目标[^3]。
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