basic calculator (stack problem)

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

 1 public class Solution {
 2     public int calculate(String s) {
 3         
 4         int len = s.length();
 5         Stack<String> stack = new Stack<String>();
 6         int i = 0;
 7         while(i < len)
 8         {
 9             if(s.charAt(i) == ' ') 
10                 i++;
11             else if (s.charAt(i) == '(' || s.charAt(i) == '+' || s.charAt(i) == '-')
12             {
13                 stack.push(s.substring(i,i+1));
14                 i++;
15             }
16             else if(s.charAt(i) >= '0' && s.charAt(i) <= '9')
17             {
18                 int start = i;
19                 while(i < len && s.charAt(i) >= '0' && s.charAt(i) <= '9') i++;
20                 stack.push(s.substring(start, i));
21             }
22             else
23             {
24                 ArrayList<String> tmp = new ArrayList<String>();
25                 while(!stack.empty())
26                 {
27                     String top = stack.pop();
28                     if(top.equals("(")) break;
29                     tmp.add(0, top);
30                 }
31                 int sumtmp = 0;
32                 sumtmp += Integer.valueOf(tmp.get(0));
33                 for(int j = 1; j < tmp.size(); j=j+2)
34                 {
35                     if(tmp.get(j)=="-")
36                         sumtmp -= Integer.valueOf(tmp.get(j+1));
37                     else
38                         sumtmp += Integer.valueOf(tmp.get(j+1));
39                 }
40                 stack.push(Integer.toString(sumtmp));
41                 i++;
42             }
43         }
44         ArrayList<String> tmp = new ArrayList<String>();
45         while(!stack.empty())
46         {
47             String top = stack.pop();
48             tmp.add(0, top);
49         }
50         int result = 0;
51         result += Integer.valueOf(tmp.get(0));
52         for(int j = 1; j < tmp.size(); j=j+2)
53         {
54             if(tmp.get(j)=="-")
55                 result -= Integer.valueOf(tmp.get(j+1));
56             else
57                 result += Integer.valueOf(tmp.get(j+1));
58         }
59         return result;
60     }
61 }

从25 到 30这段代码，i.e.,

while(!stack.empty())
 {
        String top = stack.pop();
        if(top.equals("(")) break;
        tmp.add(0, top);
 }

为什么不能写成这样

 while(stack.peek()!="(")
 {
     String top = stack.pop();
     tmp.add(0, top);
 }

写成这样就会报错

Exception in thread "main" java.util.EmptyStackException
at java.util.Stack.peek(Stack.java:102)
at helloworld.calculate(helloworld.java:43)
at helloworld.main(helloworld.java:83)

可是我能确定执行这段代码时stack是不为空的啊~