leetcode Basic Calculator

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.

计算器问题，两个栈分别存储操作数和操作符，操作符存储+ -和（，当遇到）向前回溯计算直至第一个（，代表计算完毕一个括号，当栈顶元素为+或-时代表需要进行计算，代码：

public int calculate(String s) {    int i=0;    Stack<Character> op=new Stack<>();    Stack<Integer> num=new Stack<>();    while(i<s.length()){        while(s.charAt(i)==' '){            i++;        }        if(s.charAt(i)=='+'||s.charAt(i)=='-'||s.charAt(i)=='('){            op.push(s.charAt(i));            i++;        }        else if(s.charAt(i)==')'){            while(op.peek()!='('){                char temp=op.pop();                int num1=num.pop();                int num2=num.pop();                if(temp=='+'){                    num.push(num1+num2);                }                else if(temp=='-') num.push(num2-num1);            }            op.pop();            while(!op.isEmpty()&&op.peek()!='('){                char temp=op.pop();                int num1=num.pop();                int num2=num.pop();                if(temp=='+'){                    num.push(num1+num2);                }                else if(temp=='-') num.push(num2-num1);            }            i++;        }        else{            int k=0;            while(i<s.length()&&s.charAt(i)>='0'&&s.charAt(i)<='9'){                k=k*10+(s.charAt(i)-'0');                i++;            }            if(!op.isEmpty()&&op.peek()=='+'){                op.pop();                int temp=num.pop();                num.push(temp+k);            }            else if(!op.isEmpty()&&op.peek()=='-'){                op.pop();                num.push(num.pop()-k);            }            else num.push(k);        }    }    return num.peek();}