Simple implementation of bracket matching algorithm without using stack

Just do the data structure work when there is such a book exercise, the title is as follows:

3.2 assumes that I and O are respectively represented in the stack and the stack operation, the initial and final stack of the stack are empty, the sequence of the stack and the stack can be expressed as a sequence consisting of only I and O.

(1) What is legal in the sequence shown below?

A. ioiioioo B. Iooioiio c. iiioioio D. iiiooioo

(2) Through the analysis of (1), write an algorithm to determine whether the given sequence of operations is legal. If the law returns 1, 0 is returned. (assuming that the sequence of operations determined has been deposited into an array).

The first question thinks about it, AD

The second question originally wanted to use the stack to do, but did not seem necessary ... I used an int variable count instead of the stack, and the code is as follows (C + +):

1#include <iostream>2#include <cstring>3 /*4 Name:stack operate sequence legal judge5 Copyright:guoji6 Author:guoji7 DATE:11/10/15 18:408 Description:9 */Ten  One using namespacestd; A  - BOOLJudgeChar*arr,intN) - { the     intCount =0; -      for(inti =0; I < n; i++) { -         if(Arr[i] = ='I') { -count++; +         } -         Else if(Arr[i] = ='O') { +             if(Count = =0) A                 return 0; at             Else -count--; -         } -         Else{//illegal Character -             return 0; -         } in     } -      to     return(Count = =0);  + } -   the  * intMain () $ {Panax Notoginseng     stringinput; -          theCIN >>input; Input from keyboard +      A     Char*operate =New Char[Input.length ()];  the strcpy (Operate, input.c_str ()); String to char array +      -cout <<"Result:"<< judge (operate, Input.length ()) <<Endl;  $     //Print the Judge Redult $  -     return 0;  -}

Test the data with the first question for nothing. After finishing the homework, I want to write the parentheses to match, and so on, just the second question how a bit ... Familiar with? It's a little bit like a brace match. The parentheses match another version: "I-o" matches!!! Then changed the function of the judgment, "I" with the left parenthesis "(" substitution, "O" with the right parenthesis ")" substitute, remove the illegal character judgment ...

1 BOOLJudgeChar*arr,intN)2 {3     intCount =0;4      for(inti =0; I < n; i++) {5         if(Arr[i] = ='(') {6count++;7         }8         Else if(Arr[i] = =')') {9             if(Count = =0)Ten                 return 0; One             Else Acount--; -         } -     } the      -     return(Count = =0);  -}

Write a few expressions to test a bit, can judge.

Thought analysis: Because only need to carry out the stack operation and do not involve the specific elements, the result is based on whether the stack is empty to judge, so you can use an int variable instead of the stack, with self-addition and self-subtraction instead of the stack and the stack operation, the variable is 0 to indicate that the stack is empty. But this kind of writing can only recognize one kind of parentheses (character), and how to feel not to use the stack so rigorous, to play with the homework should still be able to ...

　　

Simple implementation of bracket matching algorithm without using stack