evaluate-reverse-polish-notation

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

下面代码在牛客网没有通过，但是感觉逻辑是对的，在vs里也输入了几个例子进行了测试，不知道没有通过的原因是不是我写的太繁琐了。

思路：定义一个栈存放数字，如果遇到运算符，就取栈最上层的两个数进行运算，然后再将运算结果压入栈。

代码：

class Solution {public:    int evalRPN(vector<string> &tokens) {        if (tokens.size()<=0)            return 0;        stack<int> st;        int a = 0, b = 0, res = 0;        for (int i = 0; i < tokens.size(); ++i) {            string cur = tokens[i];            if (!isoperator(tokens[i]) && !isnumber(tokens[i]))                return 0;            if (isnumber(cur))                st.push(strtoint(cur));            else if (isoperator(cur)) {                b = st.top();//先取到的是第二个操作数                st.pop();                a = st.top();                st.pop();                res = computab(a, b, cur);                st.push(res);            }        }        return res;    }    //执行运算符的操作    int computab(int a, int b, string operate) {        if (operate[0] == '+')            return a + b;        else if (operate[0] == '-')            return a - b;        else if (operate[0] == '*')            return a * b;        else if (operate[0] == '/')            return a / b;        return 0;    }    //判断是否是运算符    bool isoperator(string x) {        if (x.size() == 1 && (x[0] == '+' || x[0] == '-' || x[0] == '*' || x[0] == '/'))            return true;        else            return false;    }    //判断是否是数字    bool isnumber(string x) {        if (x.size() <= 0)            return false;        for (int i = 0; i < x.size(); ++i) {            if (x[i] >= '0' && x[i] <= '9')                continue;            else                return false;        }        return true;    }    //把字符串里的数字转换成int    int strtoint(string x) {        if (x.size() <= 0)            return 0;        int res = x[0] - '0';        for (int i = 1; i < x.size(); ++i) {            res = res * 10 + (x[i] - '0');        }        return res;    }};

字符串里的数字转换成int的两种写法
1、一位一位转换

int strtoint(string x) {        if (x.size() <= 0)            return 0;        int res = x[0] - '0';        for (int i = 1; i < x.size(); ++i) {            res = res * 10 + (x[i] - '0');        }        return res;    }

2、用stringstream

    stringstream ss;    ss << str;    int temp;    ss >> temp;//注意这里别写成temp>>ss    cout << temp << endl;