#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 '>'; } //判断是否为运算符 bool isOp(char c) { if(c=='+'||...
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])) num.push(s[i]); else { charc=compare(op.top(),s[i]); if(c=='<') op.push(s[i]); ...
We have given an Arithmetic Expression and we have to write a program that converts the infix to postfix using stack in C. The Expression will be given in the form of a string, where alphabetic characters i.e a-z or A-Z denotes operands and operators are ( +, –, *, / ). Expres...
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'&&...
stack.Push(infixArray[i]); } } } while(stack.Count >0) { postfix[index] =stack.Pop(); index++; } returnpostfix.TakeWhile(item => item != null).ToArray(); } 看答案 如果你替换array由A.Stack<string>,你不必跟踪你在哪里index多变的。
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. ...
总体思路是这种：遇到数字的话直接输出，遇到右括号 输出左括号上面的所有元素 ，遇到左括号入栈。
Convert Infix to Postfix Notation Initially we’ve astringS which represent the expression in infix format. Now we need a characterstack. *We’ll iterate through thestringS, from left to right. Whenever we encounter an operand we’ll print it. ...
To convert infix expression to postfix expression, we will use the stack data structure. By scanning the infix expression from left to right, when we will get any operand, simply add them to the postfix form, and for the operator and parenthesis, add them in the stack maintaining the preced...
isEmpty()) { // pop and append residual operators from stack post[j] = pop(); j++; } post[j] = '\0'; // end postfix string with null character return post; } /** * @brief Self-test implementations * @returns void */ static void test() { /* check sample test case input-...