Baidu software development written test big question 3

Evaluates an expression. The known operators are only subtraction, no negative numbers, and the result is not negative. (similar to the inverse Polish method of algorithmic thinking)

Thought: As long as the numbers into the stack can be, when needed to calculate. For symbols, there are several situations:

(1) The new symbol added before the symbol stack is not signed, then directly into the stack;

(2) The new symbol is the addition and subtraction operation, then the symbols in the previous symbol stack all out to calculate, consumes 1 symbols at a time, 2 numbers, and then the resulting results are pressed into the digital stack;

(3) The new symbol is multiplication, before the symbol is also multiplication, then only need to consume a symbol on the line;

(4) The new symbol multiplication, the original symbol plus minus, then the symbol pressure stack can.

Eg. 1+4*5+2*3, defines two stacks, each with numbers and symbols, called Numstack and Sigstack.

1) The number "1" into the stack numstack;

2) because there is no symbol in the stack, the symbol "+" into the stack sigstack;

3) Digital "4" into the stack numstack;

4) "*" belongs to the multiplication, and the top of the stack is added and reduced, so directly into the stack sigstack;

5) 5 directly into the stack numstack;

6) New symbol addition and subtraction operation, means that the previous operation can be completed, the order is: 5,4 respectively out of the stack, * out of the stack, 4*5=20 into the stack numstack, and then 20,1 respectively out of the stack, + out of the stack,

1+20=21 into the stack numstack, thus numstack leaves only 21,sigstack for empty, then "+" into the stack sigstack;

7) 2 into the stack numstack;

8) New symbol *, stack top symbol +, directly into the stack;

9) 3 into the stack.

At this point all the elements have been processed, the next can consume all the symbols, the order is: 2,3 out the Stack, * out of the stack, 3*2=6 into the stack, 6,21 out the stack, + out of the stack, 21+6=27 get results.

Here is the Java implementation code:

1  Packagepretest;2 3 ImportJava.util.Scanner;4 ImportJava.util.Stack;5 6  Public classMain3 {7      Public StaticStack Sigstack =NewStack ();//For Char8      Public StaticStack Numstack =NewStack ();//for int9     Ten      Public Static voidMain (string[] args) { OneScanner in =NewScanner (system.in); A          while(In.hasnext ()) { -String str =in.nextline (); -             intresult =cal (str); the System.out.println (result); -         } -     } -      Public Static intcal (String str) { + sigstack.clear (); - numstack.clear (); +StringBuilder SB =NewStringBuilder (); A         intLen =str.length (); at         intresult= 0; -         //complete the stacking of numbers and symbols -          for(inti=0; i<len;i++){ -             Charc =Str.charat (i); -             //Digital -             if(c>= ' 0 ' && C <= ' 9 '){ in Sb.append (c); -                 if(i==len-1){ to                     intx =Integer.parseint ((sb.tostring ())); + Numstack.push (x); -                 } the}Else{ *                 intx =Integer.parseint ((sb.tostring ())); $ Numstack.push (x);Panax NotoginsengSb.replace (0, Sb.length (), ""); -                 if(sigstack.size () = = 0){ the Sigstack.push (c); +}Else{ A                     Charexisted = (Char) Sigstack.peek (); the                     //if the symbol priority of the stack has been pressed >= the current symbol C, the operation of the stack symbol is completed +                     if(c== ' + ' | | c== '-'){ - Zhengli (); $ Sigstack.push (c); $                         //When two of them are multiplication operations, -}Else if(existed = = ' * ' | | existed = = '/')){ - Tinyadjust (c); the}Else{//new symbol multiplication, old symbol plus minus - Sigstack.push (c);Wuyi                     }     the                 } -             } Wu         } -         //Calculation Results About Zhengli (); $result = (int) Numstack.pop (); -         returnresult;  -     } -      A     //Simple Subtraction Calculation +      Public Static intComputeintAintBCharc) { the         intresult = 0; -         Switch(c) { $          Case' + ': result = a +b; the              Break; the          Case'-': result = A-b; the              Break; the          Case' * ': result = A *b; -              Break; in          Case'/': result = A/b; the              Break; the         } About         returnresult; the     } the      the     //fully digest the already stacked symbol for the new symbol is the plus and minus +      Public Static voidZhengli () { -          while(Sigstack.size () > 0){ the             intA = (int) Numstack.pop ();Bayi             intB = (int) Numstack.pop (); the             Charc = (Char) Sigstack.pop (); the             intx =Compute (B,A,C); - Numstack.push (x); -         } the     } the      the     //Digest a symbol for two consecutive subtraction symbols appearing the      Public Static voidTinyadjust (Charc) { -         intX1 = (int) Numstack.pop (); the         intx2 = (int) Numstack.pop (); the         intx3 = Compute (x2, x1, (Char) Sigstack.pop ()); the Numstack.push (x3);94 Sigstack.push (c); the     } the  the}

Baidu software development written test big question 3