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("Enter the string :");
scanf("%s",a);
while(a[i]!=' '){
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]=' ';
printf("%s",b);
return 0;
}
输出<
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