栈实现中缀转后缀

这是堆栈那一节的选作实验，如下：

 

算法的核心规则

【显然，我们也可用树来实现这一过程】

c++实现转化代码如下

 

 

#include"pch.h"
#include <iostream>
#include <string>
#include <stack>
#include <map>
using namespace std;

int main()
{
    string Midsort = "a+b*(c-d)-e/f";
    string Behindsort = "";
    stack<char> stk;
    map<char, int> op;
    op['('] = 0;
    op[')'] = 0;
    op['+'] = 1;
    op['-'] = 1;
    op['*'] = 2;
    op['/'] = 2;
    string::iterator it = Midsort.begin();;//可以理解为指针哟 付凤涛桑
    while (it != Midsort.end())
    {
        if (op.count(*it))
        {
            if (*it == ')')
            {
                while (stk.top() != '(')
                {
                    Behindsort += stk.top();
                    stk.pop();
                }
                stk.pop();
            }
            else if (stk.empty() || *it == '(' || op[*it] > op[stk.top()])
            {
                stk.push(*it);
            }
            else if (op[*it] <= op[stk.top()])
            {
                while (op[*it] <= op[stk.top()] && (!stk.empty()))
                {
                    Behindsort += stk.top();
                    stk.pop();
                    if (stk.empty()) break;
                }
                stk.push(*it);
            }
        }
        else
        {
            Behindsort += *it;
        }
        it++;

        if (it == Midsort.end())
        {
            while (!stk.empty())
            {
                Behindsort += stk.top();
                stk.pop();
            }
            break;
        }
    }
    cout << Behindsort << endl;
    return 0;
}

@付风涛