Cycle sort on paikallaan oleva epävakaa lajittelualgoritmi, joka on erityisen hyödyllinen lajitettaessa taulukoita, jotka sisältävät elementtejä, joilla on pieni arvoalue. Sen on kehittänyt W. D. Jones ja se julkaistiin vuonna 1963.

Syklilajittelun perusideana on jakaa syöttötaulukko sykleihin, joissa jokainen sykli koostuu elementeistä, jotka kuuluvat samaan paikkaan lajitetussa lähtötaulukossa. Algoritmi suorittaa sitten sarjan vaihtoja sijoittaakseen jokaisen elementin oikeaan paikkaansa sen syklissä, kunnes kaikki syklit ovat valmiit ja matriisi on lajiteltu.

Tässä on vaiheittainen selitys jakson lajittelualgoritmista:

1. Aloita lajittelemattomalla n elementin joukolla.
2. Alusta muuttujan sykliAloita arvoon 0.
3. Vertaa jokaista taulukon elementtiä jokaiseen sen oikealla puolella olevaan elementtiin. Jos elementtejä on pienempi kuin nykyinen elementin lisäys cycleStart.
4. Jos cycleStart on edelleen 0 ensimmäisen elementin ja kaikkien muiden elementtien vertailun jälkeen, siirry seuraavaan elementtiin ja toista vaihe 3.
5. Kun pienempi elementti löytyy, vaihda nykyinen elementti syklin ensimmäisen elementin kanssa. Jaksoa jatketaan sitten, kunnes nykyinen elementti palaa alkuperäiseen paikkaansa.

Toista vaiheet 3-5, kunnes kaikki syklit on suoritettu.

Taulukko on nyt lajiteltu.

Yksi syklilajittelun eduista on, että sillä on pieni muistitilavuus, koska se lajittelee taulukon paikoilleen eikä vaadi lisämuistia väliaikaisia ​​muuttujia tai puskureita varten. Se voi kuitenkin olla hidasta tietyissä tilanteissa, varsinkin kun syöttötaulukolla on suuri arvoalue. Siitä huolimatta syklilajittelu on edelleen hyödyllinen lajittelualgoritmi tietyissä yhteyksissä, kuten lajitettaessa pieniä matriiseja, joilla on rajoitettu arvoalue.

Kiertolajittelu on paikallaan tapahtuva lajittelualgoritmi epävakaa lajittelualgoritmi ja vertailulajittelu, joka on teoreettisesti optimaalinen alkuperäiseen taulukkoon tehtyjen kirjoitusten kokonaismäärän suhteen. 
 

• Se on optimaalinen muistikirjoitusten lukumäärän suhteen. Se minimoi muistin kirjoitusten määrän lajitella (Jokainen arvo kirjoitetaan joko nolla kertaa, jos se on jo oikeassa paikassa tai kirjoitetaan kerran oikeaan paikkaansa.)
• Se perustuu ajatukseen, että lajiteltava taulukko voidaan jakaa sykleihin. Syklit voidaan visualisoida kaaviona. Meillä on n solmua ja reuna, joka on suunnattu solmusta i solmuun j, jos i:nnen indeksin elementin on oltava lajitellun taulukon j:nnessä indeksissä. 
  Kierto saapuessa[] = {2 4 5 1 3} 
   

• Kierto saapuessa[] = {4 3 2 1} 
   

Käsittelemme yksitellen kaikkia syklejä. Tarkastellaan ensin sykliä, joka sisältää ensimmäisen elementin. Etsimme ensimmäisen elementin oikean sijainnin ja asetamme sen oikeaan kohtaan sanomalla j. Tarkastelemme arr[j]:n vanhaa arvoa ja etsimme sen oikean paikan, jatkamme tätä, kunnes kaikki nykyisen syklin elementit ovat sijoittuneet oikeaan paikkaan, eli emme palaa syklin aloituspisteeseen.

Pseudokoodi:

 Begin   
 for   
 start:= 0 to n - 2 do   
 key := array[start]   
 location := start   
 for i:= start + 1 to n-1 do   
  if array[i]  < key then   
  location: =location +1   
 done   
 if location = start then   
  ignore lower part go for next iteration   
 while key = array[location] do   
  location: = location + 1   
 done   
 if location != start then   
  swap array[location] with key   
 while location != start do   
  location start   
  
  
 for i:= start + 1 to n-1 do   
  if array[i]  < key then   
  location: =location +1   
 done   
 while key= array[location]   
  location := location +1   
  if key != array[location]   
  Swap array[location] and key   
  done   
  done   
 End  
    Selitys:       
 
  arr[] = {10 5 2 3}   
  index = 0 1 2 3   
 cycle_start = 0    
    item     = 10 = arr[0]   
  
 Find position where we put the item    
 pos = cycle_start   
 i=pos+1   
 while(i   
 if (arr[i]  < item)    
  pos++;   
  
 We put 10 at arr[3] and change item to    
 old value of arr[3].   
 arr[] = {10 5 2        10      }    
    item     = 3    
  
 Again rotate rest cycle that start with      index '0'      
 Find position where we put the item = 3    
 we swap item with element at arr[1] now    
 arr[] = {10        3       2        10      }    
    item     = 5   
  
 Again rotate rest cycle that start with index '0' and item = 5    
 we swap item with element at arr[2].   
 arr[] = {10        3              5              10       }    
    item     = 2   
  
 Again rotate rest cycle that start with index '0' and item = 2   
 arr[] = {       2              3                      5              10      }    
  
 Above is one iteration for cycle_stat = 0.   
 Repeat above steps for cycle_start = 1 2 ..n-2  
 Alla on yllä olevan lähestymistavan toteutus:  
 CPP   
   // C++ program to implement cycle sort   #include          using     namespace     std  ;   // Function sort the array using Cycle sort   void     cycleSort  (  int     arr  []     int     n  )   {      // count number of memory writes      int     writes     =     0  ;      // traverse array elements and put it to on      // the right place      for     (  int     cycle_start     =     0  ;     cycle_start      <=     n     -     2  ;     cycle_start  ++  )     {      // initialize item as starting point      int     item     =     arr  [  cycle_start  ];      // Find position where we put the item. We basically      // count all smaller elements on right side of item.      int     pos     =     cycle_start  ;      for     (  int     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos  ++  ;      // If item is already in correct position      if     (  pos     ==     cycle_start  )      continue  ;      // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ])      pos     +=     1  ;      // put the item to it's right position      if     (  pos     !=     cycle_start  )     {      swap  (  item       arr  [  pos  ]);      writes  ++  ;      }      // Rotate rest of the cycle      while     (  pos     !=     cycle_start  )     {      pos     =     cycle_start  ;      // Find position where we put the element      for     (  int     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos     +=     1  ;      // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ])      pos     +=     1  ;      // put the item to it's right position      if     (  item     !=     arr  [  pos  ])     {      swap  (  item       arr  [  pos  ]);      writes  ++  ;      }      }      }      // Number of memory writes or swaps      // cout  < < writes  < < endl ;   }   // Driver program to test above function   int     main  ()   {      int     arr  []     =     {     1       8       3       9       10       10       2       4     };      int     n     =     sizeof  (  arr  )     /     sizeof  (  arr  [  0  ]);      cycleSort  (  arr       n  );      cout      < <     'After sort : '      < <     endl  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      cout      < <     arr  [  i  ]      < <     ' '  ;      return     0  ;   }   
  Java   
   // Java program to implement cycle sort   import     java.util.*  ;   import     java.lang.*  ;   class   GFG     {      // Function sort the array using Cycle sort      public     static     void     cycleSort  (  int     arr  []       int     n  )      {      // count number of memory writes      int     writes     =     0  ;      // traverse array elements and put it to on      // the right place      for     (  int     cycle_start     =     0  ;     cycle_start      <=     n     -     2  ;     cycle_start  ++  )     {      // initialize item as starting point      int     item     =     arr  [  cycle_start  ]  ;      // Find position where we put the item. We basically      // count all smaller elements on right side of item.      int     pos     =     cycle_start  ;      for     (  int     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos  ++  ;      // If item is already in correct position      if     (  pos     ==     cycle_start  )      continue  ;      // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ]  )      pos     +=     1  ;      // put the item to it's right position      if     (  pos     !=     cycle_start  )     {      int     temp     =     item  ;      item     =     arr  [  pos  ]  ;      arr  [  pos  ]     =     temp  ;      writes  ++  ;      }      // Rotate rest of the cycle      while     (  pos     !=     cycle_start  )     {      pos     =     cycle_start  ;      // Find position where we put the element      for     (  int     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos     +=     1  ;      // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ]  )      pos     +=     1  ;      // put the item to it's right position      if     (  item     !=     arr  [  pos  ]  )     {      int     temp     =     item  ;      item     =     arr  [  pos  ]  ;      arr  [  pos  ]     =     temp  ;      writes  ++  ;      }      }      }      }      // Driver program to test above function      public     static     void     main  (  String  []     args  )      {      int     arr  []     =     {     1       8       3       9       10       10       2       4     };      int     n     =     arr  .  length  ;      cycleSort  (  arr       n  );      System  .  out  .  println  (  'After sort : '  );      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      System  .  out  .  print  (  arr  [  i  ]     +     ' '  );      }   }   // Code Contributed by Mohit Gupta_OMG  <(0_o)>   
  Python3   
   # Python program to implement cycle sort   def   cycleSort  (  array  ):   writes   =   0   # Loop through the array to find cycles to rotate.   for   cycleStart   in   range  (  0     len  (  array  )   -   1  ):   item   =   array  [  cycleStart  ]   # Find where to put the item.   pos   =   cycleStart   for   i   in   range  (  cycleStart   +   1     len  (  array  )):   if   array  [  i  ]    <   item  :   pos   +=   1   # If the item is already there this is not a cycle.   if   pos   ==   cycleStart  :   continue   # Otherwise put the item there or right after any duplicates.   while   item   ==   array  [  pos  ]:   pos   +=   1   array  [  pos  ]   item   =   item     array  [  pos  ]   writes   +=   1   # Rotate the rest of the cycle.   while   pos   !=   cycleStart  :   # Find where to put the item.   pos   =   cycleStart   for   i   in   range  (  cycleStart   +   1     len  (  array  )):   if   array  [  i  ]    <   item  :   pos   +=   1   # Put the item there or right after any duplicates.   while   item   ==   array  [  pos  ]:   pos   +=   1   array  [  pos  ]   item   =   item     array  [  pos  ]   writes   +=   1   return   writes   # driver code    arr   =   [  1     8     3     9     10     10     2     4   ]   n   =   len  (  arr  )   cycleSort  (  arr  )   print  (  'After sort : '  )   for   i   in   range  (  0     n  )   :   print  (  arr  [  i  ]   end   =   ' '  )   # Code Contributed by Mohit Gupta_OMG  <(0_o)>   
  C#   
   // C# program to implement cycle sort   using     System  ;   class     GFG     {          // Function sort the array using Cycle sort      public     static     void     cycleSort  (  int  []     arr       int     n  )      {      // count number of memory writes      int     writes     =     0  ;      // traverse array elements and       // put it to on the right place      for     (  int     cycle_start     =     0  ;     cycle_start      <=     n     -     2  ;     cycle_start  ++  )      {      // initialize item as starting point      int     item     =     arr  [  cycle_start  ];      // Find position where we put the item.       // We basically count all smaller elements       // on right side of item.      int     pos     =     cycle_start  ;      for     (  int     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos  ++  ;      // If item is already in correct position      if     (  pos     ==     cycle_start  )      continue  ;      // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ])      pos     +=     1  ;      // put the item to it's right position      if     (  pos     !=     cycle_start  )     {      int     temp     =     item  ;      item     =     arr  [  pos  ];      arr  [  pos  ]     =     temp  ;      writes  ++  ;      }      // Rotate rest of the cycle      while     (  pos     !=     cycle_start  )     {      pos     =     cycle_start  ;      // Find position where we put the element      for     (  int     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos     +=     1  ;      // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ])      pos     +=     1  ;      // put the item to it's right position      if     (  item     !=     arr  [  pos  ])     {      int     temp     =     item  ;      item     =     arr  [  pos  ];      arr  [  pos  ]     =     temp  ;      writes  ++  ;      }      }      }      }      // Driver program to test above function      public     static     void     Main  ()      {      int  []     arr     =     {     1       8       3       9       10       10       2       4     };      int     n     =     arr  .  Length  ;          // Function calling      cycleSort  (  arr       n  );      Console  .  WriteLine  (  'After sort : '  );      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      Console  .  Write  (  arr  [  i  ]     +     ' '  );      }   }   // This code is contributed by Nitin Mittal   
  JavaScript   
    <  script  >   // Javascript program to implement cycle sort      // Function sort the array using Cycle sort      function     cycleSort  (  arr       n  )      {          // count number of memory writes      let     writes     =     0  ;          // traverse array elements and put it to on      // the right place      for     (  let     cycle_start     =     0  ;     cycle_start      <=     n     -     2  ;     cycle_start  ++  )      {          // initialize item as starting point      let     item     =     arr  [  cycle_start  ];          // Find position where we put the item. We basically      // count all smaller elements on right side of item.      let     pos     =     cycle_start  ;      for     (  let     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos  ++  ;          // If item is already in correct position      if     (  pos     ==     cycle_start  )      continue  ;          // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ])      pos     +=     1  ;          // put the item to it's right position      if     (  pos     !=     cycle_start  )      {      let     temp     =     item  ;      item     =     arr  [  pos  ];      arr  [  pos  ]     =     temp  ;      writes  ++  ;      }          // Rotate rest of the cycle      while     (  pos     !=     cycle_start  )      {      pos     =     cycle_start  ;          // Find position where we put the element      for     (  let     i     =     cycle_start     +     1  ;     i      <     n  ;     i  ++  )      if     (  arr  [  i  ]      <     item  )      pos     +=     1  ;          // ignore all duplicate elements      while     (  item     ==     arr  [  pos  ])      pos     +=     1  ;          // put the item to it's right position      if     (  item     !=     arr  [  pos  ])     {      let     temp     =     item  ;      item     =     arr  [  pos  ];      arr  [  pos  ]     =     temp  ;      writes  ++  ;      }      }      }      }       // Driver code       let     arr     =     [     1       8       3       9       10       10       2       4     ];      let     n     =     arr  .  length  ;      cycleSort  (  arr       n  );          document  .  write  (  'After sort : '     +     '  
'  );      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )      document  .  write  (  arr  [  i  ]     +     ' '  );          // This code is contributed by susmitakundugoaldanga.    <  /script>   
    
  Lähtö   
After sort : 1 2 3 4 8 9 10 10  
 
    Aika monimutkaisuusanalyysi    :   
 
 
    Huonoin tapaus:    O(n   2   )   
 
    Keskimääräinen tapaus:    O(n   2   )   
 
    Paras tapaus:    O(n   2   )  
 
 
    Aputila:    O(1)  
 
 
 Tilan monimutkaisuus on vakio, koska tämä algoritmi on paikallaan, joten se ei käytä ylimääräistä muistia lajitteluun.  
 
 
    Tapa 2:    Tätä menetelmää voidaan soveltaa vain, kun tietyt taulukon arvot tai elementit ovat välillä 1 - N tai 0 - N. Tässä menetelmässä taulukkoa ei tarvitse kiertää  
 
    Lähestymistapa:    Kaikkien annettujen taulukon arvojen tulee olla välillä 1 - N tai 0 - N. Jos alue on 1 - N, jokaisen taulukon elementin oikea sijainti on indeksi == arvo-1 eli keskiarvo 0. indeksin arvossa on 1, samoin 1. indeksin sijainnin arvo on 2 ja niin edelleen n:nteen arvoon asti.  
 
 samoin arvoille 0 - N kunkin taulukon elementin tai arvon oikea indeksisijainti on sama kuin sen arvo, eli 0. indeksissä 0 on siellä 1. paikka 1.  
 
    Selitys:     
 
 arr[] = {5 3 1 4 2}   
 index = 0 1 2 3 4   
  
 i = 0;   
 while( i  < arr.length)   
  correctposition = arr[i]-1;    
     
  find ith item correct position   
  for the first time i = 0 arr[0] = 5 correct index of 5 is 4 so arr[i] - 1 = 5-1 = 4   
     
     
  if( arr[i]  <= arr.length && arr[i] != arr[correctposition])   
     
     
  arr[i] = 5 and arr[correctposition] = 4    
  so 5  <= 5 && 5 != 4 if condition true    
  now swap the 5 with 4    
     
     
  int temp = arr[i];   
  arr[i] = arr[correctposition];   
  arr[correctposition] = temp;   
     
  now resultant arr at this after 1st swap    
  arr[] = {2 3 1 4 5} now 5 is shifted at its correct position    
     
  now loop will run again check for i = 0 now arr[i] is = 2    
  after swapping 2 at its correct position    
  arr[] = {3 2 1 4 5}   
     
  now loop will run again check for i = 0 now arr[i] is = 3    
  after swapping 3 at its correct position    
  arr[] = {1 2 3 4 5}   
     
  now loop will run again check for i = 0 now arr[i] is = 1   
  this time 1 is at its correct position so else block will execute and i will increment i = 1;   
  once i exceeds the size of array will get array sorted.   
  arr[] = {1 2 3 4 5}   
     
     
  else   
     
  i++;   
 loop end;   
  
 once while loop end we get sorted array just print it    
 for( index = 0 ; index  < arr.length; index++)   
  print(arr[index] + ' ')   
 sorted arr[] = {1 2 3 4 5}  
 Alla on yllä olevan lähestymistavan toteutus:  
 C++   
   #include          using     namespace     std  ;   void     cyclicSort  (  int     arr  []     int     n  ){      int     i     =     0  ;         while  (  i      <     n  )      {      // as array is of 1 based indexing so the      // correct position or index number of each      // element is element-1 i.e. 1 will be at 0th      // index similarly 2 correct index will 1 so      // on...      int     correct     =     arr  [  i  ]     -     1     ;      if  (  arr  [  i  ]     !=     arr  [  correct  ]){      // if array element should be lesser than      // size and array element should not be at      // its correct position then only swap with      // its correct position or index value      swap  (  arr  [  i  ]     arr  [  correct  ])     ;      }  else  {      // if element is at its correct position      // just increment i and check for remaining      // array elements      i  ++     ;      }      }   }   void     printArray  (  int     arr  []     int     size  )   {      int     i  ;      for     (  i     =     0  ;     i      <     size  ;     i  ++  )      cout      < <     arr  [  i  ]      < <     ' '  ;      cout      < <     endl  ;   }   int     main  ()     {      int     arr  []     =     {     3       2       4       5       1  };      int     n     =     sizeof  (  arr  )     /     sizeof  (  arr  [  0  ]);      cout      < <     'Before sorting array:   n  '  ;      printArray  (  arr       n  );      cyclicSort  (  arr       n  );      cout      < <     'Sorted array:   n  '  ;      printArray  (  arr       n  );      return     0  ;   }   
  Java   
   // java program to check implement cycle sort   import     java.util.*  ;   public     class   MissingNumber     {      public     static     void     main  (  String  []     args  )      {      int  []     arr     =     {     3       2       4       5       1     };      int     n     =     arr  .  length  ;      System  .  out  .  println  (  'Before sort :'  );      System  .  out  .  println  (  Arrays  .  toString  (  arr  ));      CycleSort  (  arr       n  );          }      static     void     CycleSort  (  int  []     arr       int     n  )      {      int     i     =     0  ;      while     (  i      <     n  )     {      // as array is of 1 based indexing so the      // correct position or index number of each      // element is element-1 i.e. 1 will be at 0th      // index similarly 2 correct index will 1 so      // on...      int     correctpos     =     arr  [  i  ]     -     1  ;      if     (  arr  [  i  ]      <     n     &&     arr  [  i  ]     !=     arr  [  correctpos  ]  )     {      // if array element should be lesser than      // size and array element should not be at      // its correct position then only swap with      // its correct position or index value      swap  (  arr       i       correctpos  );      }      else     {      // if element is at its correct position      // just increment i and check for remaining      // array elements      i  ++  ;      }      }      System  .  out  .  println  (  'After sort : '  );      System  .  out  .  print  (  Arrays  .  toString  (  arr  ));              }      static     void     swap  (  int  []     arr       int     i       int     correctpos  )      {      // swap elements with their correct indexes      int     temp     =     arr  [  i  ]  ;      arr  [  i  ]     =     arr  [  correctpos  ]  ;      arr  [  correctpos  ]     =     temp  ;      }   }   // this code is contributed by devendra solunke   
  Python   
   # Python program to check implement cycle sort   def   cyclicSort  (  arr     n  ):   i   =   0   while   i    <   n  :   # as array is of 1 based indexing so the   # correct position or index number of each   # element is element-1 i.e. 1 will be at 0th   # index similarly 2 correct index will 1 so   # on...   correct   =   arr  [  i  ]   -   1   if   arr  [  i  ]   !=   arr  [  correct  ]:   # if array element should be lesser than   # size and array element should not be at   # its correct position then only swap with   # its correct position or index value   arr  [  i  ]   arr  [  correct  ]   =   arr  [  correct  ]   arr  [  i  ]   else  :   # if element is at its correct position   # just increment i and check for remaining   # array elements   i   +=   1   def   printArray  (  arr  ):   print  (  *  arr  )   arr   =   [  3     2     4     5     1  ]   n   =   len  (  arr  )   print  (  'Before sorting array:'  )   printArray  (  arr  )   # Function Call   cyclicSort  (  arr     n  )   print  (  'Sorted array:'  )   printArray  (  arr  )   # This Code is Contributed by Prasad Kandekar(prasad264)   
  C#   
   using     System  ;   public     class     GFG     {      static     void     CycleSort  (  int  []     arr       int     n  )      {      int     i     =     0  ;      while     (  i      <     n  )     {      // as array is of 1 based indexing so the      // correct position or index number of each      // element is element-1 i.e. 1 will be at 0th      // index similarly 2 correct index will 1 so      // on...      int     correctpos     =     arr  [  i  ]     -     1  ;      if     (  arr  [  i  ]      <     n     &&     arr  [  i  ]     !=     arr  [  correctpos  ])     {      // if array element should be lesser than      // size and array element should not be at      // its correct position then only swap with      // its correct position or index value      swap  (  arr       i       correctpos  );      }      else     {      // if element is at its correct position      // just increment i and check for remaining      // array elements      i  ++  ;      }      }      Console  .  Write  (  'nAfter sort : '  );      for     (  int     index     =     0  ;     index      <     n  ;     index  ++  )      Console  .  Write  (  arr  [  index  ]     +     ' '  );      }      static     void     swap  (  int  []     arr       int     i       int     correctpos  )      {      // swap elements with their correct indexes      int     temp     =     arr  [  i  ];      arr  [  i  ]     =     arr  [  correctpos  ];      arr  [  correctpos  ]     =     temp  ;      }      static     public     void     Main  ()      {      // Code      int  []     arr     =     {     3       2       4       5       1     };      int     n     =     arr  .  Length  ;      Console  .  Write  (  'Before sort : '  );      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      Console  .  Write  (  arr  [  i  ]     +     ' '  );      CycleSort  (  arr       n  );      }   }   // This code is contributed by devendra solunke   
  JavaScript   
   // JavaScript code for the above code   function     cyclicSort  (  arr       n  )     {      var     i     =     0  ;      while     (  i      <     n  )      {          // as array is of 1 based indexing so the      // correct position or index number of each      // element is element-1 i.e. 1 will be at 0th      // index similarly 2 correct index will 1 so      // on...      let     correct     =     arr  [  i  ]     -     1  ;      if     (  arr  [  i  ]     !==     arr  [  correct  ])      {          // if array element should be lesser than      // size and array element should not be at      // its correct position then only swap with      // its correct position or index value      [  arr  [  i  ]     arr  [  correct  ]]     =     [  arr  [  correct  ]     arr  [  i  ]];      }      else     {      // if element is at its correct position      // just increment i and check for remaining      // array elements      i  ++  ;      }      }   }   function     printArray  (  arr       size  )     {      for     (  var     i     =     0  ;     i      <     size  ;     i  ++  )     {      console  .  log  (  arr  [  i  ]     +     ' '  );      }      console  .  log  (  'n'  );   }   var     arr     =     [  3       2       4       5       1  ];   var     n     =     arr  .  length  ;   console  .  log  (  'Before sorting array: n'  );   printArray  (  arr       n  );   cyclicSort  (  arr       n  );   console  .  log  (  'Sorted array: n'  );   printArray  (  arr       n  );   // This Code is Contributed by Prasad Kandekar(prasad264)   
    
  Lähtö   
Before sorting array: 3 2 4 5 1 Sorted array: 1 2 3 4 5  
 
    Aika monimutkaisuusanalyysi:    
 
 
    Huonoin tapaus:    O(n)   
 
    Keskimääräinen tapaus:    O(n)   
 
    Paras tapaus:    O(n)  
 
 
    Aputila:    O(1)  
 
    Cycle-lajittelun etu:    
 
 
 Lisätallennustilaa ei tarvita.  
 
  paikan päällä oleva lajittelualgoritmi.  
 
  Vähimmäismäärä kirjoituksia muistiin  
 
  Kiertolajittelu on hyödyllinen, kun taulukko on tallennettu EEPROM- tai FLASH-muistiin.   
 
 
    Cycle-lajittelun haittapuoli:    
 
 
  Sitä ei enimmäkseen käytetä.  
 
  Sillä on enemmän aikamonimutkaisuutta o(n^2)  
 
  Epävakaa lajittelualgoritmi.  
 
 
    Cycle-lajittelun sovellus:    
 
 
 Tämä lajittelualgoritmi sopii parhaiten tilanteisiin, joissa muistin kirjoitus- tai vaihtotoiminnot ovat kalliita.  
 
 Hyödyllinen monimutkaisiin ongelmiin.    
    
 
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