infix expression 计算完全版

#include<iostream>
#include<stack>
#include<string>
using namespace std;

char compare(char tp, char op)
{
	if (((tp == '+' || tp == '-') && (op == '*' || op == '/')) || (tp == '#'))
		return '<';
	else if (tp == '('&&op != ')')
		return '<';
	else if ((tp == '*' || tp == '/'||tp=='+'||tp=='-') && (op == '('))
		return '<';
	return '>';
}

int compute(int n1, int n2, char op)
{
	if (op == '+')
		return n1 + n2;
	else if (op == '-')
		return n1 - n2;
	else if (op == '*')
		return n1*n2;
	else if (op == '/')
		return n2 / n1;
}

int main()
{
	stack<char>num;
	stack<char>oper;

	oper.push('#');

	string s;
	cin >> s;

	for (int i = 0; i<s.length(); i++)
	{
		if (s[i] == '0' || s[i] == '1' || s[i] == '2' || s[i] == '3' || s[i] == '4' || s[i] == '5' || s[i] == '6' || s[i] == '7' || s[i] == '8' || s[i] == '9')
			num.push(s[i]);
		else
		{
			char comp = compare(oper.top(), s[i]);
			if (comp == '<')
				oper.push(s[i]);
			else if (comp == '>')
			{
				if (s[i] == ')')
				{
					if (num.size() != 1)
					{
						while (oper.top() != '(')
						{
							int num1 = num.top() - '0';
							num.pop();
							int num2 = num.top() - '0';
							num.pop();
							int result = compute(num1, num2, oper.top());
							num.push(result + '0');
							oper.pop();
						}
					}
					oper.pop();
				}
				else
				{
					int num1 = num.top();
					num.pop();
					int num2 = num.top();
					num.pop();
					char result = compute(num1, num2, oper.top());
					num.push(result);
					oper.pop();
					oper.push(s[i]);
				}
			}
		}
	}

	if (num.size() != 1)
	{
		while (oper.top() != '#')
		{
			int num1 = num.top()-'0';
			num.pop();
			int num2 = num.top()-'0';
			num.pop();
			int result = compute(num1, num2, oper.top());
			num.push(result+'0');
			oper.pop();
		}
	}

	cout << num.top() << endl;

	return 0;
}

  

posted @ 2016-10-20 10:34  KennyRom  阅读(275)  评论(0编辑  收藏  举报