括号匹配判断

问题描述：
检查字符串表达式中的括号是否匹配；
左括号数目同有括号数目不相等即为不匹配；
去除多余的左括号或者右括号，优先保留先出现的括号；
匹配后去除无效的括号：如：((表达式)) 应为(表达式)；
只考虑小括号，不考虑先出现右括号的情况；
要求实现函数： (字符串最长长度为60；表达式正确性不需要考虑)
void Bracket(char* src, char* dst);

如果匹配则通过dst输出原串；

如果不匹配则根据要求去处多于括号后通过dst输出匹配后的串；

示例
输入：12+(345*25-34)  输出：12+(345*25-34)

输入：12+(345*(25-34)  输出： 12+(345*25-34)

输入：(12+345)*25)-34  输出： (12+345)*25-34

输入：(543+(256-43)*203))+24

                        输出：(543+(256-43)*203)+24

输入：((1+2)*((34-2))+((2*8-1)

                        输出：((1+2)*(34-2)+2*8-1)


void Bracket(char* src, char* dst);只能在这个函数中实现,不允许用栈。不允许另写函数并加以调用。src是输入的字符串，dst是输出的字符串！

 

以下是我的代码，请大家多多指点！

//Assumption:
/*
 1. there are only number(0,...9) and math operation(+,-,*,/), no other alphabet and '$'
 2. redundant expression is just like ((expression)), no (((expression)))

 3. the expression is correct, including length less than 60 and the first bracket is '('

*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define  MAX_LENGTH 60

 

void Bracket(char* src, char* dst)
{
 char BracketNum[60];
 int LeftBktNum=0, RightBktNum=0;
 char* tmpIn = src;
 char* pHead = BracketNum;
 char* pTail = BracketNum+59;
 
 int i,j,jj;
 bool bRedundant = 0;
 //1.count the number of bracket
 while(*tmpIn)
 {
  if( *tmpIn == '(' )
  {
   LeftBktNum ++;
   *pHead++ = tmpIn - src;
  }
  else if( *tmpIn == ')' )
  {
   RightBktNum ++;
   *pTail-- = tmpIn - src;
  }

  tmpIn ++;
 }
 //2.delete unmatch bracket
 if( LeftBktNum > RightBktNum )
 {
  do
  {
   pHead --;
   src[*pHead] = '$'; //the flag indicate to be deleted bracket
 
   LeftBktNum --;
  }while(LeftBktNum > RightBktNum);

 }
 else if( LeftBktNum < RightBktNum )
 {
  do
  {
   pTail ++;
   src[*pTail] = '$'; //the flag indicate to be deleted bracket
 
   RightBktNum --;
  }while(LeftBktNum < RightBktNum);
 }

 //3.delete redundant bracket
 for( i=1,j=0; i<LeftBktNum; i++)
 {
  if( BracketNum[i] - BracketNum[i-1] == 1 )
  {
   //record left bracket
   bRedundant = 1;
   BracketNum[LeftBktNum+j] = BracketNum[i];
   if( i == LeftBktNum-1 ) BracketNum[LeftBktNum+j+1] = 60;
   else BracketNum[LeftBktNum+j+1] = BracketNum[i+1];
   j+=2;
  }
 }
 if(bRedundant)
 {  
  for( i=59; i>( 60 - RightBktNum ); i--)
  {
   if( BracketNum[i-1] - BracketNum[i] == 1 )
   {
    //record right bracket 
    jj = 0;
    while(jj<j)
    {
     if( BracketNum[LeftBktNum+jj] < BracketNum[i] && BracketNum[i] < BracketNum[LeftBktNum+jj+1] )
     {      
      src[ *(BracketNum+LeftBktNum+jj) ] = '$';      
      src[ *(BracketNum+i) ] = '$';

      if( BracketNum[LeftBktNum+jj-1] == BracketNum[LeftBktNum+jj]) BracketNum[LeftBktNum+jj-1] = BracketNum[LeftBktNum+jj+1];
     }

     jj += 2;
    }

   }
  }
 }

 //4.write src into dst
 while (*src)
 {
  if ('$'!=*src)
  {
   *dst++=*src;
  }
  src++;
 }

 *dst='/0';
}

 

void main()
{
 char strin[MAX_LENGTH] = "((1+2)*((34+(3-2))+(((9*4)))+((2*8-1)";///"((1+2)*((34+(3-2))+9*4)+((2*8-1)";
 char strout[MAX_LENGTH];
 memset(strout,0,sizeof(strout));
 
 printf("the input  is %s/n",strin);
 Bracket(strin,strout);
 printf("the output is %s/n",strout);
 getchar();
}