Basic Calculator leetcode

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

因为只有加减法 我们只要考虑括号的问题, 如果括号前面是+号, 括号内运算并不改变, 之间做加减法并入结果中, 如果括号前面是-号, 那么括号内加法, 结果就要减去相应的数. 所以我们用sign来记录数字前面的加减号, stack内部存括号外面的加减号, 这样每次运算只需要num*sign*stack.peek()
用stack和一个sign来记录符号,遇到加号 sign为1, 遇到减号sign 为-1, 遇到左括号 计算当前的符号(sign* stack.peek()) 压入栈内, 遇到右括号 出栈, 遇到数字把数字* sign* stack.peek()加入结果中

public class Solution {
    public int calculate(String s) {
        Stack<Integer> stack = new Stack<Integer>();
        stack.push(1);// 先压入一个1进栈，可以理解为有个大括号在最外面

        s = s.replace(" ", "");
        int sign = 1, res = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '+') {
                sign = 1;
            } else if (c == '-') {
                sign = -1;
            } else if (c == '(') {
                stack.push(sign * stack.peek());
                sign = 1;
            } else if (c == ')') {
                stack.pop();
            } else {
                int num = 0;
                while (i < s.length() && Character.isDigit(s.charAt(i))) {
                    num = num * 10 + s.charAt(i) - '0';
                    i++;
                }
                res += num * sign * stack.peek();
                i--;
            }
        }
        return res;
    }
}