Evaluate Reverse Polish Notation (STRING-TYPE CONVERTION)

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 = 0; 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 = 0;
        for(int i = 0; i < str.length(); i++)
        {
            ret = ret*10 + 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 = 0; 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 = 0;
        for(int i = 0; i < str.length(); i++)
        {
            ret = ret*10 + (str[i]-'0');
        }
        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 = 0; 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 = 0;
        bool negFlag = false;
        for(int i = 0; i < str.length(); i++)
        {
            if(str[i]=='-') negFlag = true;
            else if(negFlag)
            {
                ret = ret*10 - (str[i]-'0');
            }
            else
            {
                ret = ret*10 + (str[i]-'0');
            }
        }
        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 = 0; 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