Tulosta kaikki n-numeroiset luvut parittomien ja parillisten numeroiden erolla 1

Annettu kokonaisluku n joka edustaa numeroiden määrää. Tehtävänä on tulostaa kaikki n-numeroisia lukuja siten, että parillisten ja parittomien paikkojen numeroiden summan välinen absoluuttinen ero on täsmälleen 1 .
Huom : Numeron ei pitäisi alkaa (etunollia ei sallita).

Esimerkkejä:  

Syöte : n = 2
Lähtö : 10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98

Syöte : n = 3
Lähtö : 100 111 120 122 131 133 142 144 153 155 164 166 175 177 186
188 197 199 210 221 230 232 241 243 252 254 263 265 274 276 285
287 296 298 320 331 340 342 351 353 362 364 373 375 384 386 395
397 430 441 450 452 461 463 472 474 483 485 494 496 540 551 560
562 571 573 582 584 593 595 650 661 670 672 681 683 692 694 760
771 780 782 791 793 870 881 890 892 980 991  

[Odotettu lähestymistapa] Rekursion käyttö

Ideana on rekursiivisesti generoi kaikki n-numeroiset luvut samalla summan seuraaminen numeroista klo jopa ja outoa kahdella muuttujalla. Tietyn paikan kohdalla täytämme sen kaikilla numeroilla 0-9 ja sen perusteella, onko nykyinen sijainti parillinen vai pariton, lisäämme parillista tai paritonta summaa. Käsittelemme nollan alussa olevat tapaukset erikseen, koska niitä ei lasketa numeroiksi.
Olemme noudattaneet nollapohjaista numerointia, kuten taulukkoindeksiä. eli johtavan (vasemman puolen) numeron katsotaan olevan parillisessa paikassa ja sen vieressä olevan numeron katsotaan olevan parittomassa paikassa ja niin edelleen.

C++
   // C++ program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   #include          using     namespace     std  ;   // Recursive function to generate numbers   void     findNDigitNumsUtil  (  int     pos       int     n       int     num        int     evenSum       int     oddSum        vector   <  int  >     &  res  )     {      // If number is formed      if     (  pos     ==     n  )     {      // Check absolute difference condition      if     (  abs  (  evenSum     -     oddSum  )     ==     1  )     {      res  .  push_back  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  int     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ==     0     &&     d     ==     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ==     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }   }   // Function to prepare and collect valid numbers   vector   <  int  >     findNDigitNums  (  int     n  )     {          vector   <  int  >     res  ;      findNDigitNumsUtil  (  0       n       0       0       0       res  );          return     res  ;   }   // Driver code   int     main  ()     {      int     n     =     2  ;      vector   <  int  >     res     =     findNDigitNums  (  n  );      for     (  int     i     =     0  ;     i      <     res  .  size  ();     i  ++  )     {      cout      < <     res  [  i  ]      < <     ' '  ;      }      return     0  ;   }   
Java
   // Java program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   import     java.util.*  ;   class   GfG     {      // Recursive function to generate numbers      static     void     findNDigitNumsUtil  (  int     pos       int     n       int     num        int     evenSum       int     oddSum        ArrayList   <  Integer  >     res  )     {      // If number is formed      if     (  pos     ==     n  )     {      // Check absolute difference condition      if     (  Math  .  abs  (  evenSum     -     oddSum  )     ==     1  )     {      res  .  add  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  int     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ==     0     &&     d     ==     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ==     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }      }      // Function to prepare and collect valid numbers      static     ArrayList   <  Integer  >     findNDigitNums  (  int     n  )     {      ArrayList   <  Integer  >     res     =     new     ArrayList   <>  ();      findNDigitNumsUtil  (  0       n       0       0       0       res  );      return     res  ;      }      // Driver code      public     static     void     main  (  String  []     args  )     {      int     n     =     2  ;      ArrayList   <  Integer  >     res     =     findNDigitNums  (  n  );      // Print all collected valid numbers      for     (  int     i     =     0  ;     i      <     res  .  size  ();     i  ++  )     {      System  .  out  .  print  (  res  .  get  (  i  )     +     ' '  );      }      }   }   
Python
   # Python program to print all n-digit numbers such that   # the absolute difference between the sum of digits at   # even and odd positions is 1   # Recursive function to generate numbers   def   findNDigitNumsUtil  (  pos     n     num     evenSum     oddSum     res  ):   # If number is formed   if   pos   ==   n  :   # Check absolute difference condition   if   abs  (  evenSum   -   oddSum  )   ==   1  :   res  .  append  (  num  )   return   # Digits to consider at current position   for   d   in   range  (  10  ):   # Skip leading 0   if   pos   ==   0   and   d   ==   0  :   continue   # If position is even (0-based) add to evenSum   if   pos   %   2   ==   0  :   findNDigitNumsUtil  (  pos   +   1     n     num   *   10   +   d     evenSum   +   d     oddSum     res  )   # If position is odd add to oddSum   else  :   findNDigitNumsUtil  (  pos   +   1     n     num   *   10   +   d     evenSum     oddSum   +   d     res  )   # Function to prepare and collect valid numbers   def   findNDigitNums  (  n  ):   res   =   []   findNDigitNumsUtil  (  0     n     0     0     0     res  )   return   res   # Driver code   if   __name__   ==   '__main__'  :   n   =   2   res   =   findNDigitNums  (  n  )   # Print all collected valid numbers   for   i   in   range  (  len  (  res  )):   print  (  res  [  i  ]   end  =  ' '  )   
C#
   // C# program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   using     System  ;   using     System.Collections.Generic  ;   class     GfG     {      // Recursive function to generate numbers      static     void     findNDigitNumsUtil  (  int     pos       int     n       int     num        int     evenSum       int     oddSum        List   <  int  >     res  )     {      // If number is formed      if     (  pos     ==     n  )     {      // Check absolute difference condition      if     (  Math  .  Abs  (  evenSum     -     oddSum  )     ==     1  )     {      res  .  Add  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  int     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ==     0     &&     d     ==     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ==     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }      }      // Function to prepare and collect valid numbers      static     List   <  int  >     findNDigitNums  (  int     n  )     {      List   <  int  >     res     =     new     List   <  int  >  ();      findNDigitNumsUtil  (  0       n       0       0       0       res  );      return     res  ;      }      // Driver code      public     static     void     Main  (  string  []     args  )     {      int     n     =     2  ;      List   <  int  >     res     =     findNDigitNums  (  n  );      // Print all collected valid numbers      for     (  int     i     =     0  ;     i      <     res  .  Count  ;     i  ++  )     {      Console  .  Write  (  res  [  i  ]     +     ' '  );      }      }   }   
JavaScript
   // JavaScript program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   // Recursive function to generate numbers   function     findNDigitNumsUtil  (  pos       n       num       evenSum       oddSum       res  )     {      // If number is formed      if     (  pos     ===     n  )     {      // Check absolute difference condition      if     (  Math  .  abs  (  evenSum     -     oddSum  )     ===     1  )     {      res  .  push  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  let     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ===     0     &&     d     ===     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ===     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }   }   // Function to prepare and collect valid numbers   function     findNDigitNums  (  n  )     {      let     res     =     [];      findNDigitNumsUtil  (  0       n       0       0       0       res  );      return     res  ;   }   // Driver code   let     n     =     2  ;   let     res     =     findNDigitNums  (  n  );   // Print all collected valid numbers   for     (  let     i     =     0  ;     i      <     res  .  length  ;     i  ++  )     {      process  .  stdout  .  write  (  res  [  i  ]     +     ' '  );   }   

Lähtö
10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98  

Aika monimutkaisuus: O(9 × 10^(n-1)), koska jokaisessa numerossa on enintään 10 vaihtoehtoa (paitsi ensimmäisessä, jossa on 9).
Tilan monimutkaisuus: O(n + k) missä n on rekursion syvyys ja k on kelvollisten tulosten lukumäärä.