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.
 public int calPoints(String[] ops) {
if(ops == null || ops.length == 0)
return 0;
int []points = new int[ops.length];
int point = -1;
int sum = 0;
for(String op : ops){
if(op.equals("+")){
sum = points[point] + points[point - 1];
points[++ point] = sum;
}else if(op.equals("D")){
sum = points[point] * 2;
points[++ point] = sum;
}else if(op.equals("C")){
point --;
}else{
points[++point] = Integer.parseInt(op);
}
}
sum = 0;
for(int i = 0; i <= point; i++){
sum += points[i];
}
return sum;
}

优化,少遍历一遍

 public int calPoints(String[] ops) {
if(ops == null || ops.length == 0)
return 0;
int []points = new int[ops.length];
int point = -1;
int sum = 0;
int temp = 0;
for(String op : ops){
if(op.equals("+")){
temp = points[point] + points[point - 1];
points[++point] = temp;
sum += temp;
}else if(op.equals("D")){
temp = points[point] * 2;
points[++ point] = temp;
sum += temp;
}else if(op.equals("C")){
sum -= points[point];
point --;
}else{
points[++point] = Integer.parseInt(op);
sum += points[point];
}
}
return sum;
}

Baseball Game的更多相关文章

  1. Simulation of empirical Bayesian methods (using baseball statistics)

    Previously in this series: The beta distribution Empirical Bayes estimation Credible intervals The B ...

  2. [LeetCode] Baseball Game 棒球游戏

    You're now a baseball game point recorder. Given a list of strings, each string can be one of the 4 ...

  3. [Swift]LeetCode682. 棒球比赛 | Baseball Game

    You're now a baseball game point recorder. Given a list of strings, each string can be one of the 4 ...

  4. (string stoi 栈)leetcode682. Baseball Game

    You're now a baseball game point recorder. Given a list of strings, each string can be one of the 4 ...

  5. LeetCode 682 Baseball Game 解题报告

    题目要求 You're now a baseball game point recorder. Given a list of strings, each string can be one of t ...

  6. LeetCode - Baseball Game

    You're now a baseball game point recorder. Given a list of strings, each string can be one of the 4 ...

  7. [LeetCode&Python] Problem 682. Baseball Game

    You're now a baseball game point recorder. Given a list of strings, each string can be one of the 4 ...

  8. Stack-682. Baseball Game

    You're now a baseball game point recorder. Given a list of strings, each string can be one of the 4 ...

  9. Programming Assignment 3: Baseball Elimination

    编程作业三 作业链接:Baseball Elimination & Checklist 我的代码:BaseballElimination.java 问题简介 这是一个最大流模型的实际应用问题: ...

  10. 682. Baseball Game 棒球游戏 按字母处理

    [抄题]: You're now a baseball game point recorder. Given a list of strings, each string can be one of ...

随机推荐

  1. loj #107. 维护全序集

    #107. 维护全序集 题目描述 这是一道模板题,其数据比「普通平衡树」更强. 如未特别说明,以下所有数据均为整数. 维护一个多重集 S SS ,初始为空,有以下几种操作: 把 x xx 加入 S S ...

  2. loj#6041. 「雅礼集训 2017 Day7」事情的相似度(后缀自动机+启发式合并)

    题面 传送门 题解 为什么成天有人想搞些大新闻 这里写的是\(yyb\)巨巨说的启发式合并的做法(虽然\(LCT\)的做法不知道比它快到哪里去了--) 建出\(SAM\),那么两个前缀的最长公共后缀就 ...

  3. Centos查看端口占用令

    Centos查看端口占用情况命令,比如查看80端口占用情况使用如下命令: lsof -i tcp:80 列出所有端口 netstat -ntlp 1.开启端口(以80端口为例) 方法一: /sbin/ ...

  4. n阶行列式算法(c程序)

    #include<stdio.h> #include<math.h> #define N 100 //N比输入的阶数大即可 int main() {   int n,a[N][ ...

  5. 浅谈PHP的Public、Protected、Private三种方法的区别

    public:权限是最大的,可以内部调用,实例调用等.protected: 受保护类型,用于本类和继承类调用.private: 私有类型,只有在本类中使用. <?php error_report ...

  6. 好用的在线HTML、CSS工具

    css3剪贴路径(clip-path)在线生成工具:http://tools.jb51.net/static/api/css3path/index.html json在线解析:https://www. ...

  7. Android Studio 常用技巧

    1.在控制台输出语句方法 //在控制台输出语句 System.out.println("like"); //方式1 Log.d("002","lind ...

  8. Python web前端 02 CSS

    Python web前端 02 CSS 一.选择器 1.CSS的几种样式(CSS用来修饰.美化网页的) #建立模板 复制内容--->SETTING---> Editor -----> ...

  9. c# 委托访问listbox多线程操作

    c# 委托访问listbox多线程操作 using System;using System.Collections.Generic;using System.ComponentModel;using ...

  10. Codeforces - 38G 可持久化Treap 区间操作

    题意:\(n\)个人排队,每个人有重要度\(p\)和不要脸度\(c\),如果第\(i\)个人的重要度大于第\(i-1\)个人的重要度,那么他们之间可以交换,不要脸度-1,交换后先前的第\(i\)个人也 ...