Basic Calculator II

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

思路如下:
1.一次扫描，遇到+号或者-号，就将前面的数加入到结果中
2.遇到*或者/，表示这个数要与后面的数进行乘或者除

class Solution {
public:
    int calculate(string s) {
        int sign = 1, i = 0, num = getNextNum(s, i), res = 0;

        while(i < (int)s.size()){
            if(s[i] == '+'){
                res += num*sign;
                sign = 1;
                num = getNextNum(s, ++i);
            }else if(s[i] == '-'){
                res += num*sign;
                sign = -1;
                num = getNextNum(s, ++i);
            }else if(s[i] == '*'){
                num *= getNextNum(s, ++i);
            }else{
                num /= getNextNum(s, ++i);
            }

        }
        res += num*sign;
        return res;
    }
private:
    int getNextNum(string &s, int &i){
        int next = 0;
        while(i < (int)s.size()){
            if(isdigit(s[i])){
                next = next*10 + s[i] - '0';
            }else if(s[i] != ' '){
                return next;
            }
            ++i;
        }
        return next;
    }
};

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