Infix to postfix 用stack模板,表达式没有括号 #include<stack> #include<iostream> #include<string> using namespace std; //优先级判断 char compare(char opt, char si) { if((opt=='+'||opt=='-')&&(si=='*'||si=='/') return '<'; else if(opt=='#') return '<'; return '>'; }...
if(c=='+'||c=='-'||c=='*'||c=='/') returntrue; returnfalse; } intmain() { stack<char>op; stack<char>num; op.push('#'); num.push('#'); string s; cin>>s; for(inti=0;i<s.size();i++) { if(!isOp(s[i])) ...
string infix="a*(b+c+d)"; string postfix=infixToPostfix(infix); cout<<"Infix expression : "<<infix<<endl; cout<<"Postfix expression : "<<postfix<<endl; return 0; } Output: Infix Expression : a*(b+c+d) Postfix Expression : abc+d+* Time Complexity: O(n), where n is the ...
Data Structure Stack: Infix to Postfix 1#include <iostream>2#include <vector>3#include <algorithm>4#include <queue>5#include <stack>6#include <string>7#include <fstream>8#include 9#include <set>10usingnamespacestd;1112boolisoprand(charx) {13returnx >='A'&& x <='Z'|| x >='a'&&...
* @param expression - infix expre * @return postfix expr */ std::stringinfixToPostfix(std::stringexpression); //check if c is an operator boolisOperator(charc) { if(c=='+'||c=='-'||c=='*'||c=='/'||c=='^'){ returntrue; ...
I have written a C++ program to convert an infix expression to postfix expression using recursion. I would like to know if it can be improved if possible. Can we improve it by not usingastack? I am using avectorvector<char>as a stack here. ...
DATA STRUCTURES Application of Stack – Infix to Postfix conversion a Joshua Presentation A+B^C*D$ ABC^D*+ # # + # + ^ # + # + * # + # Infix Form Post fix Form Stack Infix to Postfix conversion Infix to Postfix conversion - Example 1 A+B*C^D*E+F ABCD^*E*+F+ Infix Form...
System.out.println(c.calculat()); // 5 } } 中缀转化成后缀表达式 package Stack.Calculate; import java.util.*; public class InfixToPostfix { private String infixExp = null; private List<String> postfixExp = new ArrayList<String>();
(3a - b) / 4 + c (0.5 ab) / 100 (n + 1) / n x(7x + 5) Convert the following postfix expressions to infix: a b - b a + / a b 10 * / Quizzie 2-a answers Infix to postfix: 1. a b + c * or c a b + * ...
[opstack.peek()]>=prec[token]):postfixList.append(opstack.pop())opstack.push(token)whilenotopstack.isEmpty():postfixList.append(opstack.pop())return" ".join(postfixList)# print(infixToPostfix("A * B + C * D"))print(infixToPostfix("( A + B ) * C - ( D - E ) * ( F ...