Alternating split of a given Singly Linked List | Set 1 - GeeksforGeeks

04 Apr, 2022
Write a function AlternatingSplit() that takes one list and divides up its nodes to make two smaller lists ‘a’ and ‘b’. The sublists should be made from alternating elements in the original list. So if the original list is 0->1->0->1->0->1 then one sublist should be 0->0->0 and the other should be 1->1->1.  
Method 1(Simple) The simplest approach iterates over the source list and pull nodes off the source and alternately put them at the front (or beginning) of ‘a’ and b’. The only strange part is that the nodes will be in the reverse order that occurred in the source list. Method 2 inserts the node at the end by keeping track of the last node in sublists.  
C++
C
Python
Javascript
/* C++ Program to alternatively splita linked list into two halves */#include <bits/stdc++.h>using namespace std; /* Link list node */class Node{    public:    int data;    Node* next;}; /* pull off the front node ofthe source and put it in dest */void MoveNode(Node** destRef, Node** sourceRef) ; /* Given the source list, split itsnodes into two shorter lists. If we numberthe elements 0, 1, 2, ... then all the evenelements should go in the first list, andall the odd elements in the second. Theelements in the new lists may be in any order. */void AlternatingSplit(Node* source, Node** aRef,                            Node** bRef){    /* split the nodes of source    to these 'a' and 'b' lists */    Node* a = NULL;    Node* b = NULL;             Node* current = source;    while (current != NULL)    {        MoveNode(&a, ¤t); /* Move a node to list 'a' */        if (current != NULL)        {            MoveNode(&b, ¤t); /* Move a node to list 'b' */        }    }    *aRef = a;    *bRef = b;} /* Take the node from the front ofthe source, and move it to the frontof the dest. It is an error to callthis with the source list empty.     Before calling MoveNode():source == {1, 2, 3}dest == {1, 2, 3}         After calling MoveNode():source == {2, 3}    dest == {1, 1, 2, 3}    */void MoveNode(Node** destRef, Node** sourceRef){    /* the front source node */    Node* newNode = *sourceRef;    assert(newNode != NULL);             /* Advance the source pointer */    *sourceRef = newNode->next;             /* Link the old dest off the new node */    newNode->next = *destRef;             /* Move dest to point to the new node */    *destRef = newNode;} /* UTILITY FUNCTIONS *//* Function to insert a node atthe beginning of the linked list */void push(Node** head_ref, int new_data){    /* allocate node */    Node* new_node = new Node();         /* put in the data */    new_node->data = new_data;         /* link the old list off the new node */    new_node->next = (*head_ref);             /* move the head to point to the new node */    (*head_ref) = new_node;} /* Function to print nodesin a given linked list */void printList(Node *node){    while(node!=NULL)    {    cout<<node->data<<" ";    node = node->next;    }} /* Driver code*/int main(){    /* Start with the empty list */    Node* head = NULL;    Node* a = NULL;    Node* b = NULL;         /* Let us create a sorted linked list to test the functions    Created linked list will be 0->1->2->3->4->5 */    push(&head, 5);    push(&head, 4);    push(&head, 3);    push(&head, 2);    push(&head, 1);                                    push(&head, 0);         cout<<"Original linked List: ";    printList(head);         /* Remove duplicates from linked list */    AlternatingSplit(head, &a, &b);         cout<<"\nResultant Linked List 'a' : ";    printList(a);                 cout<<"\nResultant Linked List 'b' : ";    printList(b);                 return 0;} // This code is contributed by rathbhupendra
/*Program to alternatively split a linked list into two halves */#include<stdio.h>#include<stdlib.h>#include<assert.h> /* Link list node */struct Node{    int data;    struct Node* next;}; /* pull off the front node of the source and put it in dest */void MoveNode(struct Node** destRef, struct Node** sourceRef) ; /* Given the source list, split its nodes into two shorter lists.  If we number the elements 0, 1, 2, ... then all the even elements  should go in the first list, and all the odd elements in the second.  The elements in the new lists may be in any order. */void AlternatingSplit(struct Node* source, struct Node** aRef,                            struct Node** bRef){  /* split the nodes of source to these 'a' and 'b' lists */  struct Node* a = NULL;  struct Node* b = NULL;     struct Node* current = source;  while (current != NULL)  {    MoveNode(&a, ¤t); /* Move a node to list 'a' */    if (current != NULL)    {       MoveNode(&b, ¤t); /* Move a node to list 'b' */    }  }  *aRef = a;  *bRef = b;} /* Take the node from the front of the source, and move it to the front of the dest.   It is an error to call this with the source list empty.       Before calling MoveNode():   source == {1, 2, 3}     dest == {1, 2, 3}          After calling MoveNode():   source == {2, 3}        dest == {1, 1, 2, 3}     */void MoveNode(struct Node** destRef, struct Node** sourceRef){  /* the front source node  */  struct Node* newNode = *sourceRef;  assert(newNode != NULL);     /* Advance the source pointer */  *sourceRef = newNode->next;     /* Link the old dest off the new node */  newNode->next = *destRef;     /* Move dest to point to the new node */  *destRef = newNode;} /* UTILITY FUNCTIONS *//* Function to insert a node at the beginning of the linked list */void push(struct node** head_ref, int new_data){  /* allocate node */  struct Node* new_node =            (struct Node*) malloc(sizeof(struct Node));   /* put in the data  */  new_node->data  = new_data;   /* link the old list off the new node */  new_node->next = (*head_ref);       /* move the head to point to the new node */  (*head_ref)    = new_node;} /* Function to print nodes in a given linked list */void printList(struct Node *node){  while(node!=NULL)  {   printf("%d ", node->data);   node = node->next;  }} /* Driver program to test above functions*/int main(){  /* Start with the empty list */  struct Node* head = NULL;  struct Node* a = NULL;  struct Node* b = NULL;    /* Let us create a sorted linked list to test the functions   Created linked list will be 0->1->2->3->4->5 */  push(&head, 5);  push(&head, 4);  push(&head, 3);  push(&head, 2);  push(&head, 1);                                     push(&head, 0);    printf("\n Original linked List:  ");  printList(head);   /* Remove duplicates from linked list */  AlternatingSplit(head, &a, &b);   printf("\n Resultant Linked List 'a' ");  printList(a);              printf("\n Resultant Linked List 'b' ");  printList(b);              getchar();  return 0;}
# Python program to alternatively split# a linked list into two halves # Node classclass Node:         def __init__(self, data, next = None):                 self.data = data        self.next = None class LinkedList:         def __init__(self):                 self.head = None         # Given the source list, split its    # nodes into two shorter lists. If we number    # the elements 0, 1, 2, ... then all the even    # elements should go in the first list, and    # all the odd elements in the second. The    # elements in the new lists may be in any order.    def AlternatingSplit(self, a, b):                 first = self.head        second = first.next                 while (first is not None and              second is not None and          first.next is not None):                             # Move a node to list 'a'              self.MoveNode(a, first)                             # Move a node to list 'b'              self.MoveNode(b, second)                             first = first.next.next                             if first is None:                break                             second = first.next                 # Pull off the front node of the    # source and put it in dest    def MoveNode(self, dest, node):                 # Make the new node        new_node = Node(node.data)                 if dest.head is None:            dest.head = new_node        else:                         # Link the old dest off the new node            new_node.next = dest.head                         # Move dest to point to the new node            dest.head = new_node     # UTILITY FUNCTIONS    # Function to insert a node at     # the beginning of the linked list    def push(self, data):                 # 1 & 2 allocate the Node &        # put the data                 new_node = Node(data)                 # Make the next of new Node as head        new_node.next = self.head                 # Move the head to point to new Node        self.head = new_node             # Function to print nodes    # in a given linked list    def printList(self):                 temp = self.head        while temp:            print temp.data,            temp = temp.next                     print("") # Driver Codeif __name__ == "__main__":         # Start with empty list    llist = LinkedList()    a = LinkedList()    b = LinkedList()         # Created linked list will be    # 0->1->2->3->4->5    llist.push(5)    llist.push(4)    llist.push(3)    llist.push(2)    llist.push(1)    llist.push(0)         llist.AlternatingSplit(a, b)         print "Original Linked List: ",    llist.printList()         print "Resultant Linked List 'a' : ",    a.printList()         print "Resultant Linked List 'b' : ",    b.printList()     # This code is contributed by kevalshah5
<script>// JavaScript program to alternatively split// a linked list into two halves // Node classclass Node{         constructor(data,next = null){        this.data = data        this.next = next    }} class LinkedList{       constructor()    {           this.head = null    }         // Given the source list, split its    // nodes into two shorter lists. If we number    // the elements 0, 1, 2, ... then all the even    // elements should go in the first list, and    // all the odd elements in the second. The    // elements in the new lists may be in any order.    AlternatingSplit(a, b){                 let first = this.head        let second = first.next                 while (first != null &&              second != null &&          first.next != null){                             // Move a node to list 'a'              this.MoveNode(a, first)                             // Move a node to list 'b'              this.MoveNode(b, second)                             first = first.next.next                             if(first == null)                break                             second = first.next        }    }                 // Pull off the front node of the    // source and put it in dest    MoveNode(dest, node){                 // Make the new node        let new_node = new Node(node.data)                 if(dest.head == null)            dest.head = new_node        else{                         // Link the old dest off the new node            new_node.next = dest.head                         // Move dest to point to the new node            dest.head = new_node        }    }     // UTILITY FUNCTIONS    // Function to insert a node at     // the beginning of the linked list    push(data){                 // 1 & 2 allocate the Node &        // put the data                 let new_node = new Node(data)                 // Make the next of new Node as head        new_node.next = this.head                 // Move the head to point to new Node        this.head = new_node    }             // Function to print nodes    // in a given linked list    printList(){                 let temp = this.head        while(temp){            document.write(temp.data," ");            temp = temp.next        }                     document.write("</br>")    }} // Driver Code     // Start with empty listlet llist = new LinkedList()let a = new LinkedList()let b = new LinkedList() // Created linked list will be// 0->1->2->3->4->5llist.push(5)llist.push(4)llist.push(3)llist.push(2)llist.push(1)llist.push(0) llist.AlternatingSplit(a, b) document.write("Original Linked List: ");llist.printList() document.write("Resultant Linked List 'a' : ");a.printList() document.write("Resultant Linked List 'b' : ");b.printList()     // This code is contributed by shinjanpatra </script>
Output:  
Original linked List: 0 1 2 3 4 5 
Resultant Linked List 'a' : 4 2 0 
Resultant Linked List 'b' : 5 3 1
Time Complexity: O(n) where n is a number of nodes in the given linked list.Method 2(Using Dummy Nodes) Here is an alternative approach that builds the sub-lists in the same order as the source list. The code uses temporary dummy header nodes for the ‘a’ and ‘b’ lists as they are being built. Each sublist has a “tail” pointer that points to its current last node — that way new nodes can be appended to the end of each list easily. The dummy nodes give the tail pointers something to point to initially. The dummy nodes are efficient in this case because they are temporary and allocated in the stack. Alternately, local “reference pointers” (which always point to the last pointer in the list instead of to the last node) could be used to avoid Dummy nodes.  
C++
C
Java
Python3
C#
Javascript
void AlternatingSplit(Node* source,                      Node** aRef, Node** bRef){    Node aDummy;         /* points to the last node in 'a' */    Node* aTail = &aDummy;    Node bDummy;         /* points to the last node in 'b' */    Node* bTail = &bDummy;    Node* current = source;    aDummy.next = NULL;    bDummy.next = NULL;    while (current != NULL)    {        MoveNode(&(aTail->next), ¤t); /* add at 'a' tail */        aTail = aTail->next; /* advance the 'a' tail */        if (current != NULL)        {            MoveNode(&(bTail->next), ¤t);            bTail = bTail->next;        }    }    *aRef = aDummy.next;    *bRef = bDummy.next;} // This code is contributed// by rathbhupendra
void AlternatingSplit(struct Node* source, struct Node** aRef,                            struct Node** bRef){  struct Node aDummy;  struct Node* aTail = &aDummy; /* points to the last node in 'a' */  struct Node bDummy;  struct Node* bTail = &bDummy; /* points to the last node in 'b' */  struct Node* current = source;  aDummy.next = NULL;  bDummy.next = NULL;  while (current != NULL)  {    MoveNode(&(aTail->next), ¤t); /* add at 'a' tail */    aTail = aTail->next; /* advance the 'a' tail */    if (current != NULL)    {      MoveNode(&(bTail->next), ¤t);      bTail = bTail->next;    }  }  *aRef = aDummy.next;  *bRef = bDummy.next;}
static void AlternatingSplit(Node source, Node aRef,                            Node bRef){  Node aDummy = new Node();  Node aTail = aDummy; /* points to the last node in 'a' */  Node bDummy = new Node();  Node bTail = bDummy; /* points to the last node in 'b' */  Node current = source;  aDummy.next = null;  bDummy.next = null;  while (current != null)  {    MoveNode((aTail.next), current); /* add at 'a' tail */    aTail = aTail.next; /* advance the 'a' tail */    if (current != null)    {      MoveNode((bTail.next), current);      bTail = bTail.next;    }  }  aRef = aDummy.next;  bRef = bDummy.next;} // This code is contributed by rutvik_56
def AlternatingSplit(source, aRef,  bRef):     aDummy = Node();  aTail = aDummy; ''' points to the last Node in 'a' '''  bDummy = Node();  bTail = bDummy; ''' points to the last Node in 'b' '''  current = source;  aDummy.next = None;  bDummy.next = None;  while (current != None):    MoveNode((aTail.next), current); ''' add at 'a' tail '''    aTail = aTail.next; ''' advance the 'a' tail '''    if (current != None):      MoveNode((bTail.next), current);      bTail = bTail.next;       aRef = aDummy.next;  bRef = bDummy.next; # This code is contributed by umadevi9616
static void AlternatingSplit(Node source, Node aRef,                            Node bRef){  Node aDummy = new Node();  Node aTail = aDummy; /* points to the last node in 'a' */  Node bDummy = new Node();  Node bTail = bDummy; /* points to the last node in 'b' */  Node current = source;  aDummy.next = null;  bDummy.next = null;  while (current != null)  {    MoveNode((aTail.next), current); /* add at 'a' tail */    aTail = aTail.next; /* advance the 'a' tail */    if (current != null)    {      MoveNode((bTail.next), current);      bTail = bTail.next;    }  }  aRef = aDummy.next;  bRef = bDummy.next;} // This code is contributed by pratham_76
<script>function AlternatingSplit( source,  aRef,                             bRef){  var aDummy = new Node();  var aTail = aDummy; /* points to the last node in 'a' */  var bDummy = new Node();  var bTail = bDummy; /* points to the last node in 'b' */  var current = source;  aDummy.next = null;  bDummy.next = null;  while (current != null)  {    MoveNode((aTail.next), current); /* add at 'a' tail */    aTail = aTail.next; /* advance the 'a' tail */    if (current != null)    {      MoveNode((bTail.next), current);      bTail = bTail.next;    }  }  aRef = aDummy.next;  bRef = bDummy.next;} // This code contributed by aashish1995</script>
Time Complexity: O(n) where n is number of node in the given linked list.Source: http://cslibrary.stanford.edu/105/LinkedListProblems.pdfPlease write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem. 
rathbhupendra
nidhi_biet
kevalshah5
rutvik_56
pratham76
aashish1995
simranarora5sos
umadevi9616
shinjanpatra
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