不用堆栈实现的括号匹配

括号匹配的练习时不时会在习题中遇到。为啥我没有用堆栈，说白了是没有把堆栈弄透彻，不太会用。去编写这个题其实当时是觉得有点没退路。

和公司某developer吹牛，说自己在学习C语言，经常做习题，然后把当时眼前看到的括号匹配问题发给了他，还在那里吹思路。对方一句用堆栈啊，我有点懵。于是很莽撞地回说可以不用的。接下来的问题嘛，就不需要多费唇舌了。Code拿出来就知有没有。

为了证明自己还是有半把刷子的，当天我就开始编写。而一直到隔天的晚上才完成了以下这个简陋的程序。过程中被那位大哥测试出了很多bug。现在想想都觉得不好意思。

#include <stdio.h>
#define MAX 1000

char input[MAX];
int pair (char array[], int max);


int main()
{
  int  sign = 0;
  extern char input[MAX];

 
  gets(input);    
  sign = pair (input,MAX);                
  
  
  if (sign == 0)
     printf("They are in pairs.\n");
  else if (sign == 1)
     printf ("They are not in pairs.\n");
  else
     printf("Error?\n");
 
  return 0;

}

int pair (char array[], int max)
{
  int brace = 0;
  int i = 0;
  int b = 0;
  int flag = 0;                                     //Set flag to 0, if braces are in pairs;
  
    
   for (i = 0; i < max; i++)
    {
	
	 if (array[i] =='(')
       { 
             brace++;
         
            for ( b = b + i + 1; b < max; b++)     //Under if condition, the loop goes further to find ')', then do brace-- and replace the ')' with '0';
              {
                 if (array[b] == ')')                
                 {
					brace--;
					array[b] = '0';                 //Do the replacement here;
					break;
                 }	
                 else
                    continue;                        				 
              }
       }
	  
    else if ( array[i] == ')')                   // Since the above loop will replace the first ')' it meets and breaks, we should not encounter')' here;
	        {
			  brace--;
			  break;			
			}    
    else
        continue;
    } 

  printf("brace = %d\n", brace);                //Test the brace value;
  if (brace != 0 )                               //If braces are in pairs, the integer brace should be 0 in the end;
    flag = 1;
 
 return flag;  
 
}

后来去看了堆栈思路的编写，才知道啥是强大。没见过世面就是这样。附上别人的堆栈代码。

#include<stdio.h>
#include<malloc.h>
#include<string.h>
#define STACK_INIT_SIZE 10
#define STACK_GROW_SIZE 5
#define ELEMTYPE char
#define OK 1
#define ERROR 0

typedef struct { 
    char * base;
    char * top;
    int    stacksize;
} Stack;

int InitStack(Stack *s) { /*Initialize the stack*/
    s->base=((char *) malloc(STACK_INIT_SIZE*sizeof(char)));

    if (!s->base)         /*cannot apply for the stack*/
    return ERROR;

    s->top=s->base;
    s->stacksize=STACK_INIT_SIZE;
    return OK;
}

int StackEmpty(Stack *s) {
    if (s->top==s->base) 
        return OK;
    else
        return ERROR;
}

int Push(Stack *s,char e){
    if (s->top-s->base>=s->stacksize) { 
        s->base=((char *)realloc(s->base,(s->stacksize+STACK_GROW_SIZE)*sizeof(char)));
        if (!s->base)
        return ERROR;

        s->stacksize+=STACK_GROW_SIZE;
        s->top=s->base+s->stacksize;   /*top的地址就是通过base的地址加上stacksize的地址得来的*/
    }
    *s->top++=e;
    return OK;
}

int Pop(Stack *s, char *e) { 
    if (StackEmpty(s)) 
    return ERROR;

    *e=*(--s->top);
    return OK;
}

int Comp(char a,char b) {
    if ((a=='('&&b!=')')
      ||(a=='['&&b!=']')
      ||(a=='{'&&b!='}')) {
        return ERROR;
    } 
    else
        return OK;
}

int Count(Stack *s) {
    char e[STACK_INIT_SIZE*2];
    char e1;
    int i;
 
    InitStack(s);
    printf("Please enter the content:\n");
    fgets(e,STACK_INIT_SIZE*2,stdin);
    if ('\n'==e[strlen(e)-1]) 
    e[strlen(e)-1]=0;
    printf("%s\n",e);

    for (i=0;e[i]!='\0';i++) {
        switch (e[i]) {
        case '(':
        case '[':
        case '{':
            Push(s,e[i]);
            break;
        case ')':
        case ']':
        case '}':
            if (StackEmpty(s)) {
                printf("%*s↖右括号多余\n",i+1,"");
                return(ERROR);
            } 
            else
                Pop(s,&e1);

            if (!Comp(e1,e[i])) {
                printf("%*s↖左右匹配出错\n",i+1,"");
                return(ERROR);
            }
        }
    }

    if (!StackEmpty(s)) {
        printf("%*s↖左括号多余\n",i,"");
        return(ERROR);
    } 
    else {
        printf("匹配正确\n");
        return(OK);
    }
}
int main()
{
    Stack s;
    Count(&s);
    free(s.base);
    return 0;
}

它是帮助我理解堆栈原理入门的代码。特此记录。