Shunting-yard algorithm in c++

问题

I need a function that takes an infix string (like "3 + 4 * 9"), and convert it to postfix (like "4 9 * 3 +").

I got it working until you throw in parentheses within parentheses. I've been working on it all day and can't figure out what I'm doing wrong- can someone with a fresh mind see it, maybe? I feel like I'm really close!

Thanks! Here's the code:

    string ExpressionManager::infixToPostfix(string infixExpression)
    {
cout << "itop Testing : " << infixExpression << endl;
string posnums = "0123456789";
string posops = "+-*/%(){}[]";
string onlyops = "+-/%*";
string space = " ";
string openbra = "([{";
string closebra = ")]}";
stack <string> nums;
stack <string> ops;
string output = "";

//check to make sure expression is valid

if (!(isBalanced(infixExpression)))
{
    cout << "Infix Expression isn't balanced!" << endl;
    return "invalid";
}


for (int i=0; i<infixExpression.size(); i++)
{
    if ((posnums.find(infixExpression[i])!=string::npos) || (posops.find(infixExpression[i])!=string::npos) || (space.find(infixExpression[i])!=string::npos))
    {}

    else
    {
        cout << "Invalid character " << infixExpression[i] << " found." << endl;
        return "invalid";
    }
}


int numcount = 0;
int opcount = 0;
//Numbers vs. Operators
for (int i = 0; i < infixExpression.size(); i++)
{
    if (posnums.find(infixExpression[i]) != -1)
    {

        while(infixExpression[i] != ' ')
        {
            if (i == (infixExpression.size()-1))
                break;
            i++;
        }
        numcount++;
    }

    if (onlyops.find(infixExpression[i]) != -1)
    {
        opcount++;
    }
}


if (opcount == (numcount - 1))
{
}
else
{
    cout << "Bad operators to numbers ratio!" << endl;
    return "invalid";
}

//Get rid of proceeding whitespaces.
int safety = 0;
int net = infixExpression.size();

while (infixExpression[0] == ' ')
{
    infixExpression.erase(0,1);
    safety++;

    if (safety>=net)
        break;
}

//cout << "At this point, it is " << postfixExpression << endl;

//the fun part! Set up stacks

for (int i =0; i< infixExpression.size(); i++)
{
    cout << "It gets hung up on character " << infixExpression[i] << endl;
    if(openbra.find(infixExpression[i]) != -1)
    {

        string word = "";
        word += infixExpression[i];
        ops.push(word);

        cout << "Pushing onto stack: " << word << endl;
    }
    else if(closebra.find(infixExpression[i]) != -1)
    {
            cout << "contents of remaining ops stack: "<< endl;
            stack <string> temp;
            temp = ops;
                for (int j = 0; j < temp.size(); j++)
                {
                    cout << "\t" << temp.top() << endl;
                    temp.pop();
                }

        while (openbra.find(ops.top()) == -1)
        {


            output += " " + ops.top();
            cout << "Pushing from stack: " << ops.top() << endl;
            ops.pop();
        }

        cout << "Pushing from stack: " << ops.top() << endl;
        ops.pop();

    }

    else if (posnums.find(infixExpression[i]) != -1)
    {

        string word = "";
        while (infixExpression[i] != ' ')
        {
            word += infixExpression[i];
            i++;
            if (i== infixExpression.size())
                break;
        }
        output += " " + word;

    }

    else if (onlyops.find(infixExpression[i]) != -1)
    {

        if (ops.size() == 0)
        {
            string word = "";
        word += infixExpression[i];
        ops.push(word);
        }
        else
        {
        int o1p = 0;
        int o2p = 0;

        if ((infixExpression[i] == '+') || (infixExpression[i] == '-'))
            o1p = 0;
        else
            o1p = 1;

        if ((ops.top() == "+") || (ops.top() == "-"))
            o2p = 0;
        else
            o2p = 1;

        while ((ops.size() > 0) && (o1p <= o2p))
        {
            output += " " + ops.top();
            cout << "(odd) Pushing from stack: " << ops.top() << endl;

        if ((ops.top() == "+") || (ops.top() == "-"))
            o2p = 0;
        else
            o2p = 1;

        if (ops.size() > 0)
        {
            ops.pop();
        }
        else
        {
            break;
        }
        }

        string word = "";
        word += infixExpression[i];
        ops.push(word);
        }
    }

}

while (output[0] == ' ')
{
    output.erase(0,1);
}

return output;
    }

回答1:

I personally think you have to study harder about Shunting-yard algorithm

Because you said that the output is like "4 9 * 3 +" , but what I have read about the algorithm and the stack operation, it should be (like " 9 4 * 3 +")

The important issue is that after classifying the number and operators, pop out all from the operator stack and push into the number stack with respect to the set conditions for which operator to be pop out

回答2:

My suggestion is that you go straight to the relevant Wiki pages that describe

1. Shunting Yard Algorithm
2. Reverse Polish Notation

I have implemented the Shunting Yard algorithm in both Java and C++ and found the Wiki pages excellent and a great source of help. They are in sufficient detail so as to enable you to implement the algorithm step by step in whatever programming language you prefer.

Another suggestion: become reasonably familiar with the practical use of stacks and queues, as these are used all over the place in these algorithms.

Please see this blog posting for some C++ and Java implementations of the said Shunting Yard algorithm.

It also contains a further section (in progress) should you wish to include other mathematical operators (sin, cos, log etc) and more complicated expressions and sub-expressions.

回答3:

Here is (the last version of) the solution. On some step it use Dijkstra's shunting yard algorithm (at the end of traverse() member function output_ member contains reverse polish notation form of the input_ expression, if we traverse it in right way).

来源：https://stackoverflow.com/questions/12851033/shunting-yard-algorithm-in-c