Leetcode-227.Basic Calculator II

Problem Description:
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.
You may assume that the given expression is always valid.

Some examples:
“3+2*2” = 7
” 3/2 ” = 1
” 3+5 / 2 ” = 5
Note: Do not use the eval built-in library function.

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Analysis:
First, we need to know that in this question , the operand are non-negative integers and the operators are ‘+’, ‘-‘, ‘*’, ‘/’, .
Because the ‘/’, ‘‘, have high priority. So we need to use a auxiliary stack to store the sum before ‘‘, ‘/’;

Here comes the code :

class Solution {
public:
   int calculate(string s){
    stack<int> stk;
    int sum = 0;
    int op = 0;//operand
    char sign = '+';//
    for (int i = 0; i < s.size(); ++i)
    {
        while (isdigit(s[i]))  //calculate the operand
        {
            op = op * 10 + s[i] - '0';
            ++i;
        }
        if(s[i] !=' ' ||i == s.size() - 1){ // when s[i] is sign or s is like "1  ", or could add a '+' in the rightmost.
            if (sign == '+') //**sign is previous operators**
            {
                stk.push(op);
            }
            else if (sign == '-')
            {
                stk.push(-op);
            }
            else if (sign == '*')
            {
                int t = stk.top() * op;
                stk.pop();
                stk.push(t);
                }
            else 
            {
                int t = stk.top() / op;
                stk.pop();
                stk.push(t);
            }
            sign = s[i]; // store the next operator.
            op = 0;
        }
    }
    while(!stk.empty())
    {
        sum += stk.top();
        stk.pop();
    }
    return sum;
}
};

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Another shorter version without stack.

Tricky trick is to add ‘+’ both side,make codes elegant.
use >> to skip the whitespace and easy to read a specific type number.
O(n), O(1)
Every time I got operator, if it is + or -, I can make a total of the left side; then read
the next digit and store with the sign as a temp value;
if it is * or /.just evaluate the result and store as a temp value.

 int calculate(string s) {
        char op;
        long long total=0,term=0,n;
        istringstream in('+' + s + '+');
        while(in>>op)
        {
            if(op=='+'|| op=='-')
                {
                  total+=term;
                  in>>term;
                  term*=op=='+'?1:-1;// term *= 44 - op;
                }
                else{
                    in>>n;
                    if(op=='*')
                        term*=n;
                    else 
                        term/=n;
                   }

        }
                return total;
        }