Реализации алгоритмов/Кодирование длин серий

C/C++

#include <stdio.h>
#include <string.h>

#define N 80

int main()
{
    char *smb, code[N], encode[2 * N] = "", str[N];
    int cnt = 1;
    
    fgets(code, 80, stdin);
    smb = code;
    
    for (int i = 1; i < strlen(code); i++, smb++, cnt++) if (code[i] != *smb)
      {
          sprintf(encode + strlen(encode), "%d%c", cnt, *smb);
          cnt = 0;
      }
    
    printf("%s\n", encode);
    return 0;
}

Python

def rle(src):
    result = []
    if src:
        current = src.pop(0)
        counter = 1
        for e in src:
            if e == current:
                counter += 1
            else:
                result.append((counter, current))
                current = e
                counter = 1
        result.append((counter, current))
    return result

PHP

<?php
$code = 'fafaaaaaaaaaaaaa';
$encode = '';

for ($i = 0; $i < strlen($code);$i++){
	$smb = $code[$i] ;
	$count = 1 ;
	for ($b = $i; $b < strlen($code);$b++){
		if ($code[$b + 1] != $smb) break ;
		$count++ ;

		$i++ ;
	}
	$encode .= $count . $smb ;
}
print 'Строку: ' . $code . ' удалось сжать до ' . $encode . '.<br> И мы сэкономили ' . (strlen($code) - strlen($encode)) . ' байт.' 
?>

Delphi/Pascal

function encode(s:string):string;
var i,k:integer; c:char;
begin
  Result:='';
  if s='' then exit;
  c:=s[1]; 
  k:=1;
  for i:=2 to length(s)+1 do
    if s[i]=c then inc(k) else
      begin
        if k>1 then Result:=Result+IntToStr(k);
        Result:=Result+c;
        c:=s[i];
        k:=1;
      end;
end;
 
function decode(s:string):string;
var i,j,c:integer;
    newS:string;
begin
i:=1;
while i <= length(s) do
  begin
    j:=i;
    while s[j] in ['0'..'9'] do inc(j);
    if j-i > 0 then
    begin
      for c:=1 to strtoint(copy(s,i,j-i)) do newS := newS + s[j];
      delete(s,i,j-i+1);
    end else
    begin
      newS := newS + s[i];
      inc(i);
    end;
  end;
  result:= newS;
end;

Visual Basic 6

Public Function Encode(ByVal SrcString As String) As String
    Dim I As Long, N As Long, sResult As String
    N = 1
    
    For I = 1 To Len(SrcString)
        If Not Mid(SrcString, I, 1) = Mid(SrcString, I + 1, 1) Then
            sResult = sResult & IIf(I - N + 1 > 1, CStr(I - N + 1), CStr("")) & Mid(SrcString, I, 1)
            N = I + 1
        End If
    Next
    
    Encode = sResult
End Function

Haskell

import Data.List ( group )

main = putStrLn . encode =<< getLine

-- group разбивает строку на подстроки из одинаковых элементов, foldr собирает строку.
encode = foldr (\ cs -> shows (length cs) . showChar (head cs)) "" . group

-- Линейно-рекурсивный вариант (явная реализация).
encode'    []    = []
encode' (x : xs) = go 1 x xs
  where
    go i x (y : ys)
      | x == y = go (i + 1) x ys         -- увеличение значения счётчика
      |  True  = shows i $ x : go 1 y ys -- присоединение элемента строки
    go i x _ = shows i [x]