CODEWITHSUNDEEP
c++stringalgorithmuniquestd
Zohar81
Zohar81

Reputation: 5068

Remove specific consecutive characters duplications from std::string

Perhaps anyone have an efficient way to remove consecutive duplications of specific characters preferably using built-in string operations, and without explicitly going through the string characters.

For example, when I have wildcard pattern and I'd like to remove consecutive asterisks only (*)
/aaaa/***/bbbb/ccc/aa/*****/dd --> /aaaa/*/bbbb/ccc/aa/*/dd

For all characters repetitive duplications I can use std::unique in the following manner :

str.erase( std::unique(str.begin(), str.end()), str.end());

but what about specific chars only.

Upvotes: 4

Views: 256

Answers (1)

Vlad from Moscow
Vlad from Moscow

Reputation: 311010

You can use the same algorithm std::unique with a lambda expression.

For example

#include <iostream>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>

int main()
{
    std::string s = "/aaaa/***/bbbb/ccc/aa/*****/dd"; 
    char c = '*';

    s.erase( std::unique( std::begin( s ), std::end( s ), 
                          [=]( const auto &c1, const auto &c2 ) { return c1 == c && c1 == c2; } ),
             std::end( s ) ); 

    std::cout << s << '\n';             
}

The program output is

/aaaa/*/bbbb/ccc/aa/*/dd

Or you can remove a set of duplicate characters. For example

#include <iostream>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <cstring>

int main()
{
    std::string s = "/aaaa/***/bbbb/ccc/aa/*****/dd"; 
    const char *targets = "*b";

    auto remove_chars = [=]( const auto &c1, const auto &c2 )
    {
        return strchr( targets, c1 ) && c1 == c2;
    };
    s.erase( std::unique( std::begin( s ), std::end( s ), remove_chars ), 
             std::end( s ) ); 

    std::cout << s << '\n';             
}

The program output is

/aaaa/*/b/ccc/aa/*/dd

In the last example I suppose that the character '\0' is not included in the string. Otherwise you have to add one more subexpression to the logical expression in the lambda.

Upvotes: 9

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