leetcode: Basic Calculator

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

/*Example: 7 - (6 - 5 - (4 - 3) - 2) - (1)Result : +7 - 6 + 5 + 4 - 3 + 2 - 1        The + -are signs just before '('.They are dealt with previous stack sign.        Then they are saved in the stack to help decide the signs in the following "(..)" .*/class Solution {public:    int calculate(string s) {        stack<int> t;  //previous sign just before the '('        t.push(1);    //for all the signs those are not in the parentheses        int sum = 0, temp = 0, lastSign = 1;        for (auto c : s)        {            if (c == '(') t.push(lastSign); //save the sign just before the '('            else if (c == ')') t.pop();            if (c >= '0' && c <= '9'){                temp = temp * 10 + c - '0';            }            if (c == '-' || c == '+'){                sum += lastSign*temp; //calculate the number before current sign                temp = 0;                lastSign = (c == '-' ? -1 : 1)*t.top();  //deal with the stack influencing sign            }        }        sum += lastSign*temp; //calculate the last number        return sum;    }};