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-negati ve integers and empty spaces .

Assume that the given expression was always valid.

Some Examples:

"1 + 1" = 2 "2-1 + 2" = 3 "(1+ (4+5+2)-3) + (6+8)" = 23

1  Public classSolution {2      Public intCalculate (String s) {3         4         intLen =s.length ();5stack<string> stack =NewStack<string>();6         inti = 0;7          while(I <len)8         {9             if(S.charat (i) = = ") Teni++; One             Else if(S.charat (i) = = ' (' | | | S.charat (i) = = ' + ' | | S.charat (i) = = '-') A             { -Stack.push (S.substring (i,i+1)); -i++; the             } -             Else if(S.charat (i) >= ' 0 ' && s.charat (i) <= ' 9 ') -             { -                 intStart =i; +                  while(I < len && S.charat (i) >= ' 0 ' && s.charat (i) <= ' 9 ') i++; - Stack.push (s.substring (Start, i)); +             } A             Else at             { -arraylist<string> tmp =NewArraylist<string>(); -                  while(!stack.empty ()) -                 { -String top =Stack.pop (); -                     if(Top.equals ("(")) Break; inTmp.add (0, top); -                 } to                 intsumtmp = 0; +Sumtmp + = integer.valueof (tmp.get (0)); -                  for(intj = 1; J < Tmp.size (); J=j+2) the                 { *                     if(Tmp.get (j) = = "-") $Sumtmp-= integer.valueof (Tmp.get (j+1));Panax Notoginseng                     Else -Sumtmp + = integer.valueof (Tmp.get (j+1)); the                 } + Stack.push (integer.tostring (sumtmp)); Ai++; the             } +         } -arraylist<string> tmp =NewArraylist<string>(); $          while(!stack.empty ()) $         { -String top =Stack.pop (); -Tmp.add (0, top); the         } -         intresult = 0;WuyiResult + = integer.valueof (tmp.get (0)); the          for(intj = 1; J < Tmp.size (); J=j+2) -         { Wu             if(Tmp.get (j) = = "-") -Result-= Integer.valueof (Tmp.get (j+1)); About             Else $Result + = Integer.valueof (Tmp.get (j+1)); -         } -         returnresult; -     } A}

From 25 to 30 this code, i.e,

 while (! Stack.empty ()) {        = stack.pop ();         if  Break ;        Tmp.add (0, top); }

Why can't it be written like this

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

Written like this will be an error

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)

But I can be sure that the stack is not empty when executing this code.

Basic calculator (stack problem)