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

Note: Do not use the eval built-in library function.

題目中只有+ - ( )。遍歷字符串，對于每個字符c：

• 如果是數字，則一直遍歷到非數字字符，把數字找出，并與結果相加
• 如果是+-符號，將sign設置成對應的值
• 如果是(，將rt和sign壓入棧中，重置rt和sign
• 如果是)，將sign和rt彈出棧，并計算結果

    /**
     * 計算帶括號的string的值
     * @param s
     * @return
     */
    public int calculate(String s) {
        if(s==null){
            throw  new IllegalArgumentException();
        }

        int res=0; //結果值
        int flag=1; //記錄前面的符號，加為1，減為-1

        Stack<Integer> stack=new Stack<>(); 

        for (int i=0;i<s.length();i++){
            char c=s.charAt(i);
            int val=0;
            if(Character.isDigit(c)){
                val=c-'0';
                while (i+1<s.length() && Character.isDigit(s.charAt(i+1))){
                    val=val*10+s.charAt(++i)-'0';
                }

                res+=flag*val;
            }
            else if(c=='+'){
                flag=1;
            }else if(c=='-'){
                flag=-1;
            }else if(c=='('){
                  //遇到括號先壓棧結果，再壓棧符號
                stack.push(res);
                stack.push(flag);
                res=0;
                flag=1;
            }else if(c==')'){
                res=res*stack.pop()+stack.pop();
            }
        }
        System.out.print(res);
        return res;
    }