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.
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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 −
#includeint main(){ int i=0,c=0,j=0; char a[100],b[100]; printf(" Enter the string :"); scanf("%s",a); while(a[i]!='\0'){ if((a[i]=='(') && (a[i-1]=='-')){ (c=0)?j=i:j=c; while(a[i]!=')'){ if(a[i+1]=='+') b[j++]='-'; else if(a[i+1]=='-') b[j++]='+'; else if(a[i+1]!=')') b[j++]=a[i+1]; i++; } c=j+1; } else if(a[i]=='(' && a[i-1]=='+'){ (c==0)?j=i:j=c; while(a[i]!=')'){ b[j++]=a[i+1]; i++; } j–; c=j+1; } else if(a[i]==')'){ i++; continue; } else { b[j++]=a[i]; } i++; } b[j]='\0'; 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
