Memandangkan program input C, keluarkan kurungan

问题

让我们通过删除表达式中的括号来创建一个简化的表达式。

解决方案

示例 1

Input: A string expression with bracket is as follows:
(x+y)+(z+q)
The output is as follows:
x+y+z+q

示例 2

The input is as follows:
(x-y+z)-p+q
The output is as follows:
x-y+z-p+q

Algorithm

Refer an algorithm to remove the brackets from a given input.

Step 1: Declare and read the input at runtime.

Step 2: Traverse the string.

Step 3: Copy each element of the input string into new string.

Step 4: If anyone parenthesis is encountered as an element, replace it with empty space.

Example

Following is the C program to remove the brackets from a given input −

#include<stdio.h>
int main(){
   int i=0,c=0,j=0;
   char a[100],b[100];
   printf("</p><p>Enter the string :");
   scanf("%s",a);
   while(a[i]!=&#39;\0&#39;){
      if((a[i]==&#39;(&#39;) && (a[i-1]==&#39;-&#39;)){
         (c=0)?j=i:j=c;
         while(a[i]!=&#39;)&#39;){
            if(a[i+1]==&#39;+&#39;)
               b[j++]=&#39;-&#39;;
            else if(a[i+1]==&#39;-&#39;)
               b[j++]=&#39;+&#39;;
            else if(a[i+1]!=&#39;)&#39;)
               b[j++]=a[i+1];
               i++;
         }
         c=j+1;
      }
      else if(a[i]==&#39;(&#39; && a[i-1]==&#39;+&#39;){
         (c==0)?j=i:j=c;
         while(a[i]!=&#39;)&#39;){
            b[j++]=a[i+1];
            i++;
         }
         j&ndash;;
         c=j+1;
      }
      else if(a[i]==&#39;)&#39;){
         i++;
         continue;
      } else {
         b[j++]=a[i];
      }
      i++;
   }
   b[j]=&#39;\0&#39;;
   printf("%s",b);
   return 0;
}

输出

执行上述程序时，会产生以下输出 -

Enter the string:(x+y)-z
x+y-z

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