所以,我正在编写一个程序,它接受输入(数学表达式)并将其转换为后缀表示法,然后将其打印出来给用户。当我用简单的数学表达式(即5 + 1 - 3 * 2
)执行这个程序时,它运行得很好。问题是当我包括括号时。它打印最后一行的所有数学运算符而不是它们的适当位置。我需要数学运算符在他们适当的位置。 (即5 1 + 3 - 2 *
)。有什么建议吗?
此声明5 - ( 2 * 3 ) + 5
的实际输出为5 2 3 5 * - +
我需要它看起来像5 2 3 * - 5 +
#include <iostream>
#include <stack>
#include <string>
#include <sstream>
using namespace std;
int priority(string item) //prioritizing the operators
{
if(item == "(" || item == ")")
{
return 0;
}
else if(item == "+" || item == "-")
{
return 1;
}
else if(item == "/" || item == "*")
{
return 2;
}
}
void welcome()
{
cout << "blah blah!" << endl;
}
int main()
{
welcome(); // welcome text
stack <string> myStack;
char line[256];
cin.getline( line, 256);
string exp = line;
string item;
string output = "";
istringstream iss(exp);
iss >> item;
while(iss)
{
if(item != "+" && item != "-" && item != "/" && item != "*" && item != "(" && item != ")") //If the char is a number
{
cout << item << endl;
}
else if(myStack.size() == 0) // If the stack is empty
{
myStack.push(item);
}
else if( item == "+" || item == "-" || item == "/" || item == "*") //If the char is an operator
{
if(priority(myStack.top()) < priority(item)) // the item on the stack is greater priority than the array item
{
myStack.push(item);
}
else
{
while(myStack.size() > 0 && priority(myStack.top()) >= priority(item)) //while the stack contains something, and the item on
{
string dpu = myStack.top();
myStack.pop();
output = output + dpu + " ";
}
myStack.push(item);
}
}
else if(item == "(") // left peren
{
myStack.push(item);
}
else if(item == ")") // right peren
{
while(myStack.top() != "(")
{
string dpu = myStack.top();
myStack.pop();
output = output + dpu + " ";
}
myStack.pop();
}
iss>> item;
}
while (myStack.size() > 0 )
{
string dpu = myStack.top();
output = output + dpu + " ";
myStack.pop();
}
cout<< output << endl;
}
答案 0 :(得分:1)
嗯......在你的代码中,数字总是先打印出来:
if (item != "+" && item != "-" && item != "/" && item != "*" && item != "(" && item != ")") //If the char is a number
{
cout << item << endl;
}
我不熟悉你要求的符号,但是堆栈在我看来并不是这个任务的正确数据结构...使用树似乎对我更好,因为你正在解析