Evaluate Reverse Polish Notation

最新推荐文章于 2022-03-25 16:52:08 发布

转载最新推荐文章于 2022-03-25 16:52:08 发布 · 45 阅读

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原文链接：http://www.cnblogs.com/pengpenghappy/p/3861606.html

本文介绍了一种用于计算逆波兰表达式的算法实现。该算法通过栈来处理操作数和运算符，能够正确解析并计算包含加、减、乘、除四种基本运算的操作表达式。文中提供了一个具体的代码示例，演示了如何使用该算法计算特定的逆波兰表达式。

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

int stoi(string str)
    {
        bool flag = true;
        if (str[0] == '-') flag = false;
        int ret = 0, i;
        i = (flag) ? 0 : 1;
        for (; i<str.size(); i++)
        {
            ret = ret * 10 + (str[i] - '0');
        }
        return (flag) ? ret : (-ret);
    }

    int evalRPN(vector<string> &tokens)
    {
        int ans = 0;
        stack<int> s;
        int a, b, c;
        for (int i = 0; i < tokens.size(); i++)
        {
            if (tokens[i] != "+"&&tokens[i] != "-"&&tokens[i] != "*"&&tokens[i] != "/")
            {
                string str = tokens[i];
                int temp = stoi(str);
                s.push(temp);
            }
            else
            {
                a = s.top();
                s.pop();
                b = s.top();
                s.pop();
                if (tokens[i] == "+") c = b + a;
                if (tokens[i] == "-") c = b - a;
                if (tokens[i] == "*") c = b*a;
                if (tokens[i] == "/") c = b / a;
                s.push(c);
            }
        }
        ans = s.top();
        return ans;
    };

int _tmain(int argc, _TCHAR* argv[])
{
    Solution *sln = new Solution();
    //string s = "  hello dd  ";
    //sln->reverseWords(s);
    //cout << s << endl;

    vector<string> svec;
    svec.push_back("4");
    svec.push_back("13");
    svec.push_back("5");
    svec.push_back("/");
    svec.push_back("+");

    int ret = sln->evalRPN(svec);

    cout << ret << endl;

    return 0;
}

转载于:https://www.cnblogs.com/pengpenghappy/p/3861606.html