An array of characters is given. Delete part of the text enclosed in brackets (along with brackets) [closed]

There are standard search functions for substrings in a string. You look for the first opening bracket and memorize its index, then in the same way you close the index and also memorize the index. The fact that between the indices can also be obtained by the standard copying functions of the adjustment starting and ending with a certain position. This will be the text between the brackets. Do with it what you want and repeat the algorithm. Also for convenience, you can use another line in which you will add the result of your actions. I think . option with another array is the best option. Go through the main array and check the conditions, then what satisfies add to another array

In short, I could not change this line and I went the other way. I copied the content to another line, only there is, let's say, one big question. If the line does not contain a closing parenthesis, then I have garbage, or rather garbage after the character '('.

 #include <iostream> #include <cstdio> using namespace std; int main() { setlocale(LC_ALL, "RUSSIAN"); char str[100] = "# :#($ afe)0t:(rk: f)akt", *ptr; char n_str[100] = "\0", *n_ptr; cout << "Данная строка: " << str << endl; ptr = str; while (*ptr) { n_ptr = n_str; while (*ptr != '(' && *ptr) { *n_ptr = *ptr; n_ptr++; ptr++; //игнорируем символы которые стоят между скобок //и сами скобки тоже игнорируем if (*ptr == '(') { *ptr++; while (*ptr != ')') ptr++; ptr++; //перемещаемся за за символ ')' } } } *n_ptr = '\0'; cout << "Новая строка: " << n_str << endl; return 0; } 

I think if it is intended to delete all completed sequences in brackets, then an algorithm using the stack for “recognizing” characters in brackets and a “queue” in which we place pairs of indices corresponding to the opening and matching closing brackets will be appropriate.

I write “queue” in quotation marks, because if the index of the opening bracket in a pair's queue is less than the index at its end, then the “queue” turns into a “stack” (from the side where the elements are usually inserted into it), t . A new pair of indexes (in fact, this is just the closed sequence of characters in brackets, which began earlier than the one already placed in the queue) replaces the elements previously placed in it.

We get this algorithm.

Looking through the characters of the string, when the opening bracket appears, we put its index on the stack. The closing bracket extracts from the stack the index of the paired opening and together they form a pair, which must be placed in the queue, taking into account the remark above.

Those. at the end we get an array ("queue") of pairs of character indexes, which should be. removed from the string. In fact, the deletion can be replaced by the "in-place" line rewriting, excluding characters from the pairs in this array.

(as always, the description is longer than the code and perhaps more incomprehensible -))

PS
@Ilya, you wrote that you want to write the code yourself, if it will still be interesting, scribble in the comments (indicating @avp (otherwise I will not see the notification (such is crap here))).