Run-length encoding: Difference between revisions
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'''Final note''': since the repeat counter value 0 has no meaning, it could be used as it would be 256, so extending by one the maximum number of repetitions representable with a single byte; or instead it could be used as a special marker to encode in a more efficient way (long) sequences of ''isolated characters'', e.g. "ABCDE" would be encoded as "1A1B1C1D1E"; it could be instead encoded as "05ABCDE".
=={{header|C++}}==
<lang C++>
#include <iostream>
#include <string>
#include <sstream>
#include <boost/regex.hpp>
#include <cstdlib>
std::string encode ( const std::string & ) ;
std::string decode ( const std::string & ) ;
int main( ) {
std::string to_encode ( "WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW" ) ;
std::cout << to_encode << " encoded:" << std::endl ;
std::string encoded ( encode ( to_encode ) ) ;
std::cout << encoded << std::endl ;
std::string decoded ( decode( encoded ) ) ;
std::cout << "Decoded again:\n" ;
std::cout << decoded << std::endl ;
if ( to_encode == decoded )
std::cout << "It must have worked!\n" ;
return 0 ;
}
std::string encode( const std::string & to_encode ) {
std::string::size_type found = 0 , nextfound = 0 ;
std::ostringstream oss ;
nextfound = to_encode.find_first_not_of( to_encode[ found ] , found ) ;
while ( nextfound != std::string::npos ) {
oss << nextfound - found ;
oss << to_encode[ found ] ;
found = nextfound ;
nextfound = to_encode.find_first_not_of( to_encode[ found ] , found ) ;
}
//since we must not discard the last characters we add them at the end of the string
std::string rest ( to_encode.substr( found ) ) ;//last run of characters starts at position found
oss << rest.length( ) << to_encode[ found ] ;
return oss.str( ) ;
}
std::string decode ( const std::string & to_decode ) {
boost::regex e ( "(\\d+)(\\w)" ) ;
boost::match_results<std::string::const_iterator> matches ;
std::ostringstream oss ;
std::string::const_iterator start = to_decode.begin( ) , end = to_decode.end( ) ;
while ( boost::regex_search ( start , end , matches , e ) ) {
std::string numberstring ( matches[ 1 ].first , matches[ 1 ].second ) ;
int number = atoi( numberstring.c_str( ) ) ;
std::string character ( matches[ 2 ].first , matches[ 2 ].second ) ;
for ( int i = 0 ; i < number ; i++ )
oss << character ;
start = matches[ 2 ].second ;
}
return oss.str( ) ;
}
</lang>
=={{header|Clojure}}==
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