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

别人优秀的代码：

1.这个代码简洁，其中44 - op 代表 op == '+' ? 1 : -1。

同时，>>能跳过空格

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 *= 44 - op;
        } else {
            in >> n;
            if (op == '*')
                term *= n;
            else
                term /= n;
        }
    }
    return total;
}

2.容易理解，也是很通用的好方法。也是别人优秀的代码

int calculate(string s) {
        int result = 0, cur_res = 0;
        char op = '+';
        for(int pos = s.find_first_not_of(' '); pos < s.size(); pos = s.find_first_not_of(' ', pos)) {
            if(isdigit(s[pos])) {
                int tmp = s[pos] - '0';
                while(++pos < s.size() && isdigit(s[pos]))
                    tmp = tmp*10 + (s[pos] - '0');
                switch(op) {
                    case '+' : cur_res += tmp; break;
                    case '-' : cur_res -= tmp; break;
                    case '*' : cur_res *= tmp; break;
                    case '/' : cur_res /= tmp; break;
                }
            }
            else {
                if(s[pos] == '+' || s[pos] == '-') {
                    result += cur_res;
                    cur_res = 0;
                }
                op = s[pos++];
            }
        }
        return result + cur_res;
    }

3.这个用栈的也是很简单的方法

int calculate(string s) {
    stack<int> myStack;
    char sign = '+';
    int res = 0, tmp = 0;
    for (unsigned int i = 0; i < s.size(); i++) {
        if (isdigit(s[i]))
            tmp = 10*tmp + s[i]-'0';
        if (!isdigit(s[i]) && !isspace(s[i]) || i == s.size()-1) {
            if (sign == '-')
                myStack.push(-tmp);
            else if (sign == '+')
                myStack.push(tmp);
            else {
                int num;
                if (sign == '*' )
                    num = myStack.top()*tmp;
                else
                    num = myStack.top()/tmp;
                myStack.pop();
                myStack.push(num);
            } 
            sign = s[i];
            tmp = 0;
        }
    }
    while (!myStack.empty()) {
        res += myStack.top();
        myStack.pop();
    }
    return res;
}