LeetCode OJ Evaluate Reverse Polish Notation

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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.

Some examples:

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

后缀表达式的计算，从左往右扫，数字进栈，符号就提取栈顶两个数字计算再进栈

注意只用提交类。

class Solution {public:    int evalRPN(vector<string> &tokens) {        vector<int> stack_int;        for (int i = 0; i < tokens.size(); i++) {            int x, y;            if ('0' <= tokens[i][tokens[i].size() - 1] && tokens[i][tokens[i].size() - 1] <= '9') {                stack_int.push_back(string_to_num(tokens[i]));            } else {                y = stack_int.back();                stack_int.pop_back();                x = stack_int.back();                stack_int.pop_back();                if (tokens[i][0] == '+') {                    stack_int.push_back(x + y);                } else if (tokens[i][0] == '-') {                    stack_int.push_back(x - y);                } else if (tokens[i][0] == '*') {                    stack_int.push_back(x * y);                } else if (tokens[i][0] == '/') {                    stack_int.push_back(x / y);                }            }        }        return stack_int.back();    }    int string_to_num(string input) {        int sum = 0;        int positive = 1;        int i = 0;        if (input[0] == '-') {            positive = -1;            i = 1;        }        for (; i < input.size(); i++) {            sum = sum * 10 + input[i] - '0';        }        return sum * positive;    }};

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