You're now a baseball game point recorder.

Given a list of strings, each string can be one of the 4 following types:

  1. Integer (one round's score): Directly represents the number of points you get in this round.
  2. "+" (one round's score): Represents that the points you get in this round are the sum of the last two valid round's points.
  3. "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.
  4. "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.

Just use a stack to solve this problem.

class Solution:
def calPoints(self, ops):
"""
:type ops: List[str]
:rtype: int
"""
ans=0
baseballstack=collections.deque() for c in ops:
if c!='C' and c!='D' and c!='+':
num=int(c)
ans+=num
baseballstack.appendleft(num)
elif c=='C':
num=baseballstack.popleft()
ans-=num
elif c=='D':
num=baseballstack[0]
num*=2
ans+=num
baseballstack.appendleft(num)
elif c=='+':
num=baseballstack[0]
num2=baseballstack[1]
num+=num2
ans+=num
baseballstack.appendleft(num) return ans

  

[LeetCode&Python] Problem 682. Baseball Game的更多相关文章

  1. [LeetCode&Python] Problem 108. Convert Sorted Array to Binary Search Tree

    Given an array where elements are sorted in ascending order, convert it to a height balanced BST. Fo ...

  2. [LeetCode&Python] Problem 387. First Unique Character in a String

    Given a string, find the first non-repeating character in it and return it's index. If it doesn't ex ...

  3. [LeetCode&Python] Problem 427. Construct Quad Tree

    We want to use quad trees to store an N x N boolean grid. Each cell in the grid can only be true or ...

  4. [LeetCode&Python] Problem 371. Sum of Two Integers

    Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -. Exam ...

  5. [LeetCode&Python] Problem 520. Detect Capital

    Given a word, you need to judge whether the usage of capitals in it is right or not. We define the u ...

  6. [LeetCode&Python] Problem 226. Invert Binary Tree

    Invert a binary tree. Example: Input: 4 / \ 2 7 / \ / \ 1 3 6 9 Output: 4 / \ 7 2 / \ / \ 9 6 3 1 Tr ...

  7. [LeetCode&Python] Problem 905: Sort Array By Parity

    Given an array A of non-negative integers, return an array consisting of all the even elements of A, ...

  8. [LeetCode&Python] Problem 1: Two Sum

    Problem Description: Given an array of integers, return indices of the two numbers such that they ad ...

  9. [LeetCode&Python] Problem 811. Subdomain Visit Count

    A website domain like "discuss.leetcode.com" consists of various subdomains. At the top le ...

随机推荐

  1. STL_容器使用时机

    1. 来自教程: ◆ Vector的使用场景:比如软件历史操作记录的存储,我们经常要查看历史记录,比如上一次的记录,上上次的记录,但却不会去删除记录,因为记录是事实的描述. ◆ deque的使用场景: ...

  2. Android JNI学习(三)——Java与Native相互调用

    本系列文章如下: Android JNI(一)——NDK与JNI基础 Android JNI学习(二)——实战JNI之“hello world” Android JNI学习(三)——Java与Nati ...

  3. unity项目针对IOS及Android平台的音频压缩格式

    IOS : 建议采用MP3格式, Android : 建议采用Vorbis格式, 因为这两种格式分别在这两个平台上有硬件解码的支持, 硬件解码比软件解码快.

  4. C#对实体进行JSON序列化时枚举的处理

    实体类如下: public enum ESex { Boy, Girl } public class Person { public String Name { get; set; } public ...

  5. Redis之有序集合命令

    Redis 有序集合(sorted set) Redis 有序集合和集合一样也是string类型元素的集合,且不允许重复的成员. 不同的是每个元素都会关联一个double类型的分数.redis正是通过 ...

  6. mybatis: 多对多查询[转]

    加入3个包 log4j-1.2.17.jar mybatis-3.3.0.jar mysql-connector-java-5.1.8.jar log4j需要配置 log4j.properties # ...

  7. js下载图片

    DownloadImgZP = imgPath => { const image = new Image(); // 解决跨域 Canvas 污染问题 image.setAttribute('c ...

  8. yield 关键字

    yield 关键字向编译器指示它所在的方法是迭代器块.编译器生成一个类来实现迭代器块中表示的行为.在迭代器块中,yield 关键字与 return 关键字结合使用,向枚举器对象提供值.这是一个返回值, ...

  9. EBS管理员为供应商创建新联系人流程

    管理员为供应商创建新联系人流程 /oracle/apps/pos/supplier/webui/ByrAddCntctPG oracle.apps.pos.supplier.webui.ByrAddC ...

  10. CentOS7系统更换YUM Repo源

    CentOS7系统更换YUM Repo源 备份原镜像 sudo mv /etc/yum.repos.d/CentOS-Base.repo /etc/yum.repos.d/CentOS-Base.re ...