LeetCode 150：Evaluate Reverse Polish Notation-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_34314736/article/details/100652319

本文介绍了一种算法，用于评估逆波兰表示法中的算术表达式的值。该算法使用栈来处理运算符和操作数，支持加、减、乘、除四种基本运算。文章通过三个示例详细解释了算法的运行过程。

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Evaluate the value of an arithmetic expression in Reverse Polish Notation.

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

Note:

Division between two integers should truncate toward zero.
The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation.
Example 1:

Input: [“2”, “1”, “+”, “3”, “*”]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:

Input: [“4”, “13”, “5”, “/”, “+”]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:

Input: [“10”, “6”, “9”, “3”, “+”, “-11”, “", “/”, "”, “17”, “+”, “5”, “+”]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

C++
 int evalRPN(vector<string>& tokens) {
        stack<int> s;
        int res;
        int len = tokens.size();
        for(int i = 0;i < len;++i)
        {
            if(tokens[i] != "+"&&tokens[i] != "-"&&tokens[i] != "*"&&tokens[i] != "/")
            {
                s.push(stoi(tokens[i]));
            }
            else
            {
                int second = s.top();
                s.pop();
                int first = s.top();
                s.pop();
                if(tokens[i] == "+")s.push(first+second);
                else if(tokens[i] == "-")s.push(first-second);
                else if(tokens[i] == "*")s.push(first*second);
                else if(tokens[i] == "/")s.push(first/second);
                else
                    break;
            }
        }
        res = s.top();
        return res;
    }