我已经编写了这段代码来使用 CPP 中的堆栈将中缀表达式转换为后缀表达式

Question

#include<bits/stdc++.h>
using namespace std;

int prec(char c){
    if(c=='^'){
        return 3;
    }else if(c=='*' || c=='/'){
        return 2;
    }else if(c=='+' || c=='-'){
        return 1;
    }

    return -1;
}

string infixToPostfix(string );

int main(){
    string s = "(a-b/c)*(a/k-l)";

    cout<<infixToPostfix(s);

    return 0;
}

string infixToPostfix(string s){
    stack<char> st;
    string res = "";

    for(int i=0;i<s.length();i++){
        if((s[i]>='a' && s[i]<='z') || (s[i]>='A' && s[i]<='Z')){
            res+=s[i];
        }else if(s[i]=='('){
            st.push(s[i]);
        }else if(s[i]==')'){
            while((!st.empty()) && st.top()!='('){
                res+=st.top();
                st.pop();
            }
            if(!st.empty()){
                st.pop();
            }
        }else{
            while((!st.empty()) && prec(st.top())>prec(s[i])){
                res+=st.top();
                st.pop();
            }
            st.push(s[i]);
        }

        while(!st.empty()){
            res+=st.top();
            st.pop();
        }

    }
        return res;
}

As you can see I'm trying to convert infix notation to postfix notation but the output is not coming as expected. I couldn't even find any syntax error so there's a high chance that there is some logical error.

Expected Output:

abc/-ak/l-*

Actual Output:

(a-b/c*(a/k-l

I have blown my brain off trying to find the error and I still haven't. Please help me solve the issue.

Answer 1

Define two precedence tables, called outstack for operator when they are outside the stack and instack for operator when they are inside the stack.

If any operator is left to right assosiative increase the precedence from outstack to instack . If it is right to left decrease the precedence.

Op	outstack pre	instack pre
+ -	1	2
* /	3	4
^	6	5
(	7	0
)	0	x

The program below uses this logic to convert an infix expression to postfix expression.

#include <iostream>
#include <string>
#include <stack>
#include <map>
#include <vector>

// For left to right associative operator, precedence increase from out to in
// For right to left associative operator (like ^), precedence decrease from out to in

// Outside stack precedence
std::map<char, int> precedenceOutStack {
    {'(', 7}, 
    {')', 0},
    {'*', 3}, 
    {'/', 3},
    {'+', 1}, 
    {'-', 1},
    {'^', 6},
};

// Inside stack precedence
std::map<char, int> precedenceInStack {
    {'(', 0}, 
    {'*', 4}, 
    {'/', 4},
    {'+', 2}, 
    {'-', 2},
    {'^', 5},
};

int getOutPrecedence(char c) {
    if(precedenceOutStack.count(c) > 0) {
        return precedenceOutStack[c];
    } else {
        return 0;
    }
}

int getInPrecedence(char c) {
    if(precedenceInStack.count(c) > 0) {
        return precedenceInStack[c];
    } else {
        return 0;
    }
}

std::string infixToPostfix(std::string infix) {
    std::string postfix {};
    std::stack<char> stk;

    size_t i {};
    // loop through the input string
    while(infix[i]) {
        // if its an operand add it to postfix
        if(std::isalpha(infix[i])) {
            postfix.push_back(infix[i++]);
        } else {
            if(!stk.empty()) {
                auto outPrec = getOutPrecedence(infix[i]);
                auto inPrec = getInPrecedence(stk.top());

                // check the out precedence of input char with in precedence of stack top
                // if its greater push the operator to stack
                if( outPrec > inPrec ) {
                    stk.push(infix[i++]);
                }
                // else if it is less, append the operator from top of stack to postfix
                else if(outPrec < inPrec ) {
                    postfix.push_back(stk.top());
                    stk.pop();    
                }
                // only '(' and ')' has equal out and in precedence, ignore them 
                else if(outPrec == inPrec) {
                    stk.pop();
                    ++i;
                }
            } else {
                stk.push(infix[i++]);
            }
        }
    }

    // pop out remaining opreator from the stack and append them to postfix
    while(!stk.empty()) {
        postfix.push_back(stk.top());
        stk.pop();
    }
    return postfix;
}

int main()
{
    std::vector<std::string> inputs {
        "(a-b/c)*(a/k-l)"  // abc/-ak/l-*
    };

    for(const auto& s : inputs) {
        std::cout << "Infix: " << s << " -- ";
        std::cout << infixToPostfix(s) << std::endl;
    }
}