Question

Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

1ST TRY

class Solution {
public:
int evalRPN(vector<string> &tokens) {
int operant1;
int operant2;
string str;
int ret;
bool firstOp = false; for(int i = ; i < tokens.size(); i++)
{
str = tokens[i];
if(!firstOp)
{
operant1 = getInt(str);;
firstOp = true;
}
else if(str == "+")
{
ret = operant1 + operant2;
}
else if(str == "-")
{
ret = operant1 - operant2;
}
else if(str == "*")
{
ret = operant1 * operant2;
}
else if(str == "/")
{
ret = operant1 / operant2;
}
else
{
operant2 = getInt(str);
firstOp = false;
}
} return ret;
}
int getInt(string str)
{
int ret = ;
for(int i = ; i < str.length(); i++)
{
ret = ret* + str[i];
}
return ret;
}
};

2ND TRY

考虑只有一个操作数的情况

class Solution {
public:
int evalRPN(vector<string> &tokens) {
int operant1;
int operant2;
string str;
int ret;
bool firstOp = false; for(int i = ; i < tokens.size(); i++)
{
str = tokens[i];
if(!firstOp)
{
operant1 = getInt(str);
firstOp = true;
}
else if(str == "+")
{
operant1 += operant2;
}
else if(str == "-")
{
operant1 -= operant2;
}
else if(str == "*")
{
operant1 *= operant2;
}
else if(str == "/")
{
operant1 /= operant2;
}
else
{
operant2 = getInt(str);
}
}
return operant1;
} int getInt(string str)
{
int ret = ;
for(int i = ; i < str.length(); i++)
{
ret = ret* + (str[i]-'');
}
return ret;
}
};

Result: Wrong

Input: ["3","-4","+"]
Output: -23
Expected: -1

3RD TRY

考虑负数的情况

class Solution {
public:
int evalRPN(vector<string> &tokens) {
int operant1;
int operant2;
int ret;
string str;
stack<int> operantStack; for(int i = ; i < tokens.size(); i++)
{
str = tokens[i];
if(str == "+")
{
operant2 = operantStack.top();
operantStack.pop();
operant1 = operantStack.top();
operantStack.pop();
operantStack.push(operant1 + operant2);
}
else if(str == "-")
{
operant2 = operantStack.top();
operantStack.pop();
operant1 = operantStack.top();
operantStack.pop();
operantStack.push(operant1 - operant2);
}
else if(str == "*")
{
operant2 = operantStack.top();
operantStack.pop();
operant1 = operantStack.top();
operantStack.pop();
operantStack.push(operant1 * operant2);
}
else if(str == "/")
{
operant2 = operantStack.top();
operantStack.pop();
operant1 = operantStack.top();
operantStack.pop();
operantStack.push(operant1 / operant2);
}
else
{
operantStack.push(getInt(str));
}
}
return operantStack.top();
} int getInt(string str)
{
int ret = ;
bool negFlag = false;
for(int i = ; i < str.length(); i++)
{
if(str[i]=='-') negFlag = true;
else if(negFlag)
{
ret = ret* - (str[i]-'');
}
else
{
ret = ret* + (str[i]-'');
}
}
return ret;
}
};

Result: Accepted

4TH TRY

使用内置函数atoi将string转换成int

class Solution {
public:
int evalRPN(vector< string > &tokens) {
stack< int > operandStack;
int operand1;
int operand2;
for(int i = ; i < tokens.size(); i++){
if(tokens[i]=="+"){
operand1 = operandStack.top();
operandStack.pop();
operand2 = operandStack.top();
operandStack.pop();
operand2 += operand1;
operandStack.push(operand2);
}
else if(tokens[i]=="-"){
operand1 = operandStack.top();
operandStack.pop();
operand2 = operandStack.top();
operandStack.pop();
operand2 -= operand1;
operandStack.push(operand2);
}
else if(tokens[i]=="*"){
operand1 = operandStack.top();
operandStack.pop();
operand2 = operandStack.top();
operandStack.pop();
operand2 *= operand1;
operandStack.push(operand2);
}
else if(tokens[i]=="/"){
operand1 = operandStack.top();
operandStack.pop();
operand2 = operandStack.top();
operandStack.pop();
operand2 /= operand1;
operandStack.push(operand2);
}
else{
operand1 = atoi(tokens[i].c_str());
operandStack.push(operand1);
}
}
return operandStack.top();
}
};

Result: Accepted

Evaluate Reverse Polish Notation (STRING-TYPE CONVERTION)的更多相关文章

  1. LeetCode: Reverse Words in a String:Evaluate Reverse Polish Notation

    LeetCode: Reverse Words in a String:Evaluate Reverse Polish Notation Evaluate the value of an arithm ...

  2. 【leetcode】Evaluate Reverse Polish Notation

    Evaluate Reverse Polish Notation 题目描述: Evaluate the value of an arithmetic expression in Reverse Pol ...

  3. [LintCode] Evaluate Reverse Polish Notation 计算逆波兰表达式

    Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, ...

  4. 【LeetCode练习题】Evaluate Reverse Polish Notation

    Evaluate Reverse Polish Notation Evaluate the value of an arithmetic expression in Reverse Polish No ...

  5. leetcode - [2]Evaluate Reverse Polish Notation

    Evaluate Reverse Polish Notation Total Accepted: 24595 Total Submissions: 123794My Submissions Evalu ...

  6. 【LeetCode】150. Evaluate Reverse Polish Notation

    Evaluate Reverse Polish Notation Evaluate the value of an arithmetic expression in Reverse Polish No ...

  7. LeetCode: Evaluate Reverse Polish Notation 解题报告

    Evaluate Reverse Polish Notation Evaluate the value of an arithmetic expression in Reverse Polish No ...

  8. LeetCode 150. 逆波兰表达式求值(Evaluate Reverse Polish Notation) 24

    150. 逆波兰表达式求值 150. Evaluate Reverse Polish Notation 题目描述 根据逆波兰表示法,求表达式的值. 有效的运算符包括 +, -, *, /.每个运算对象 ...

  9. 【LeetCode】150. Evaluate Reverse Polish Notation 解题报告(Python)

    [LeetCode]150. Evaluate Reverse Polish Notation 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/ ...

  10. 【刷题-LeetCode】150 Evaluate Reverse Polish Notation

    Evaluate Reverse Polish Notation Evaluate the value of an arithmetic expression in Reverse Polish No ...

随机推荐

  1. mybatis 3.2.2_环境搭建

    1.创建一个工程 utf-8 2.导入jar mybatis-3.2.2.jar 核心包 依赖包: asm-3.3.1.jar cglib-2.2.2.jar commons-logging-1.1. ...

  2. BBS-文章详情页、评论、评论树

    1.简单的实现评论功能 article_detail.html,拿到数据  由路--给视图函数--写入数据库 <p>评论内容:</p> <textarea name=&q ...

  3. Tomcat 7集群基于redis的session共享设置

    经过测试之后,发现是tomcat中redis相关jar包问题,替换jar包后A产品运行正常. tomcat/lib目录下将commons-pool2-2.1.jar.jedis-2.1.0.jar.t ...

  4. ORACLE常用操作命令

    1.ORACLE实例启动.停止 SQL>startup;  #启动ORACLE实例 SQL>shutdown immediate; #关闭ORACLE实例,常用.阻止新用户连接且阻止已连接 ...

  5. t959 unknown device 解决办法

    换机器没用 换数据线没用 最后装了Kies3,好了! -------- 更新 跟数据线也有关系 换一条三星自带的试试

  6. 自定义 mapper的实现

    json格式,要想好看直接百度,json,将字符放进去就可 一步:将mapper复制一份,名字加一个Custom自定义 二步:mpper.xml也是一样,设置里面的namespace映射关系 自定义m ...

  7. Ajax 学习 第二篇

    XMLHttpRequest发送请求 open(method,url,async) 解释 method:do/post,不区分大小写 URL:相对地址 文档地址 async:默认为TRUE 即异步 F ...

  8. 一: Introduction(介绍)

    Welcome to SQLBackupRestore.com, your online resource for SQL Server backup and recovery issues.  Th ...

  9. CDH6离线安装

    CentOS7下完全离线安装CDH6集群   CDH Linux 大数据 shareSHARE   245 浏览量 分享到微博 分享到 Twitter 分享到 QQ 本文是在CentOS7.5下进行C ...

  10. poj2480-Longge's problem-(欧拉函数)

    Longge is good at mathematics and he likes to think about hard mathematical problems which will be s ...