解法

n的使用，来判断是否有数
大步骤是逆序的

代码

import java.util.Stack;
class Solution {
    public int cal(Stack<String> stack)
    {
        int res=Integer.parseInt(stack.pop());
        boolean flag=true;
        while(!stack.isEmpty()&&!stack.peek().equals(")"))
        {
            String temp=stack.pop();
            if(temp.equals("+")){
                flag=true;
            }else if(temp.equals("-"))
            {
                flag=false;
            }else
            {
                if(flag)
                {
                    res+=Integer.parseInt(temp);
                }else
                {
                    res-=Integer.parseInt(temp);
                }
            }
        }
        if(!stack.isEmpty()) stack.pop();
        return res;
    }
    public int calculate(String s) {
        char[] chars = s.toCharArray();
        Stack<String> stack=new Stack<>();
        int curnum=0;
        int n=0;
        for(int i=chars.length-1;i>=0;i--)
        {
            if(Character.isDigit(chars[i]))
            {
                curnum=(chars[i]-'0')*(int)Math.pow(10,n)+curnum;
                n++;
            }else if(chars[i]!=' ')
            {
                if(n!=0)
                {
                    stack.push(String.valueOf(curnum));
                    curnum=0;
                    n=0;
                }
                if(chars[i]=='(')
                {
                    int res = cal(stack);
                    stack.push(String.valueOf(res));
                }else
                {
                    stack.push(String.valueOf(chars[i]));
                }
            }
        }
        int res=0;
        if(n!=0)
        stack.push(String.valueOf(curnum));
        res = cal(stack);
        return res;
    }
}