[Leetcode]Basic Calculator

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces.

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2" 2-1 + 2 " = 3"(1+(4+5+2)-3)+(6+8)" = 23

Note: Do not use the eval built-in library function.

class Solution {public:    /*algorithm: stack soluton        two stack: operand stack, and operator stack        transform it to reverse polish notation,then compute        but we can skip this step        one pass    */    void tryExecute(stack<int>&opStk,stack<int>&numStk){        if(numStk.size() < 2 || opStk.empty())return;        int op = opStk.top();        if(op =='+' || op == '-'){            int r = numStk.top();numStk.pop();            int l = numStk.top();numStk.pop();            numStk.push(op == '+'?(l+r):(l-r));            opStk.pop();        }    }    int numLength(string &s,int start){        int i = start;        for(;i < s.size() && isdigit(s[i]);i++);        return i - start;    }    int calculate(string s) {        stack<int>opStk,numStk;        int next;        for(int i = 0;i < s.size();){            char c = s[i];            next = i + 1;            if(c == '(')opStk.push(c);            else if(c == ')'){                opStk.pop();                tryExecute(opStk,numStk);            } else if(c == '+' || c == '-') opStk.push(c);            else if(isdigit(c)){                int len = numLength(s,i);                numStk.push(stoi(s.substr(i,len)));                tryExecute(opStk,numStk);                next = i + len;            }            //update i pos            i = next;        }        return numStk.top();    }};