Given a linked list and a key in it, the task is to move all occurrences of given key to end of linked list, keeping order of all other elements same.
Examples:
Input : 1 -> 2 -> 2 -> 4 -> 3 key = 2 Output : 1 -> 4 -> 3 -> 2 -> 2 Input : 6 -> 6 -> 7 -> 6 -> 3 -> 10 key = 6 Output : 7 -> 3 -> 10 -> 6 -> 6 -> 6
A simple solution is to one by one find all occurrences of given key in linked list. For every found occurrence, insert it at the end. We do it till all occurrences of given key are moved to end.
Time Complexity : O(n2)
Efficient Solution 1 : is to keep two pointers:
pCrawl => Pointer to traverse the whole list one by one.
pKey => Pointer to an occurrence of key if a key is found. Else same as pCrawl.
We start both of the above pointers from head of linked list. We move pKey only when pKey is not pointing to a key. We always move pCrawl. So when pCrawl and pKey are not same, we must have found a key which lies before pCrawl, so we swap data of pCrawl and pKey, and move pKey to next location. The loop invariant is, after swapping of data, all elements from pKey to pCrawl are keys.
Below is the implementation of this approach.
C++
// C++ program to move all occurrences of a // given key to end. #include <bits/stdc++.h> // A Linked list Node struct Node { int data; struct Node* next; }; // A urility function to create a new node. struct Node* newNode( int x) { Node* temp = new Node; temp->data = x; temp->next = NULL; } // Utility function to print the elements // in Linked list void printList(Node* head) { struct Node* temp = head; while (temp != NULL) { printf ( "%d " , temp->data); temp = temp->next; } printf ( "\n" ); } // Moves all occurrences of given key to // end of linked list. void moveToEnd(Node* head, int key) { // Keeps track of locations where key // is present. struct Node* pKey = head; // Traverse list struct Node* pCrawl = head; while (pCrawl != NULL) { // If current pointer is not same as pointer // to a key location, then we must have found // a key in linked list. We swap data of pCrawl // and pKey and move pKey to next position. if (pCrawl != pKey && pCrawl->data != key) { pKey->data = pCrawl->data; pCrawl->data = key; pKey = pKey->next; } // Find next position where key is present if (pKey->data != key) pKey = pKey->next; // Moving to next Node pCrawl = pCrawl->next; } } // Driver code int main() { Node* head = newNode(10); head->next = newNode(20); head->next->next = newNode(10); head->next->next->next = newNode(30); head->next->next->next->next = newNode(40); head->next->next->next->next->next = newNode(10); head->next->next->next->next->next->next = newNode(60); printf ( "Before moveToEnd(), the Linked list is\n" ); printList(head); int key = 10; moveToEnd(head, key); printf ( "\nAfter moveToEnd(), the Linked list is\n" ); printList(head); return 0; } |
Java
// Java program to move all occurrences of a // given key to end. class GFG { // A Linked list Node static class Node { int data; Node next; } // A urility function to create a new node. static Node newNode( int x) { Node temp = new Node(); temp.data = x; temp.next = null ; return temp; } // Utility function to print the elements // in Linked list static void printList(Node head) { Node temp = head; while (temp != null ) { System.out.printf( "%d " , temp.data); temp = temp.next; } System.out.printf( "\n" ); } // Moves all occurrences of given key to // end of linked list. static void moveToEnd(Node head, int key) { // Keeps track of locations where key // is present. Node pKey = head; // Traverse list Node pCrawl = head; while (pCrawl != null ) { // If current pointer is not same as pointer // to a key location, then we must have found // a key in linked list. We swap data of pCrawl // and pKey and move pKey to next position. if (pCrawl != pKey && pCrawl.data != key) { pKey.data = pCrawl.data; pCrawl.data = key; pKey = pKey.next; } // Find next position where key is present if (pKey.data != key) pKey = pKey.next; // Moving to next Node pCrawl = pCrawl.next; } } // Driver code public static void main(String args[]) { Node head = newNode( 10 ); head.next = newNode( 20 ); head.next.next = newNode( 10 ); head.next.next.next = newNode( 30 ); head.next.next.next.next = newNode( 40 ); head.next.next.next.next.next = newNode( 10 ); head.next.next.next.next.next.next = newNode( 60 ); System.out.printf( "Before moveToEnd(), the Linked list is\n" ); printList(head); int key = 10 ; moveToEnd(head, key); System.out.printf( "\nAfter moveToEnd(), the Linked list is\n" ); printList(head); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 program to move all occurrences of a # given key to end. # Linked List node class Node: def __init__( self , data): self .data = data self . next = None # A urility function to create a new node. def newNode(x): temp = Node( 0 ) temp.data = x temp. next = None return temp # Utility function to print the elements # in Linked list def printList( head): temp = head while (temp ! = None ) : print ( temp.data,end = " " ) temp = temp. next print () # Moves all occurrences of given key to # end of linked list. def moveToEnd(head, key): # Keeps track of locations where key # is present. pKey = head # Traverse list pCrawl = head while (pCrawl ! = None ) : # If current pointer is not same as pointer # to a key location, then we must have found # a key in linked list. We swap data of pCrawl # and pKey and move pKey to next position. if (pCrawl ! = pKey and pCrawl.data ! = key) : pKey.data = pCrawl.data pCrawl.data = key pKey = pKey. next # Find next position where key is present if (pKey.data ! = key): pKey = pKey. next # Moving to next Node pCrawl = pCrawl. next return head # Driver code head = newNode( 10 ) head. next = newNode( 20 ) head. next . next = newNode( 10 ) head. next . next . next = newNode( 30 ) head. next . next . next . next = newNode( 40 ) head. next . next . next . next . next = newNode( 10 ) head. next . next . next . next . next . next = newNode( 60 ) print ( "Before moveToEnd(), the Linked list is\n" ) printList(head) key = 10 head = moveToEnd(head, key) print ( "\nAfter moveToEnd(), the Linked list is\n" ) printList(head) # This code is contributed by Arnab Kundu |
C#
// C# program to move all occurrences of a // given key to end. using System; class GFG { // A Linked list Node public class Node { public int data; public Node next; } // A urility function to create a new node. static Node newNode( int x) { Node temp = new Node(); temp.data = x; temp.next = null ; return temp; } // Utility function to print the elements // in Linked list static void printList(Node head) { Node temp = head; while (temp != null ) { Console.Write( "{0} " , temp.data); temp = temp.next; } Console.Write( "\n" ); } // Moves all occurrences of given key to // end of linked list. static void moveToEnd(Node head, int key) { // Keeps track of locations where key // is present. Node pKey = head; // Traverse list Node pCrawl = head; while (pCrawl != null ) { // If current pointer is not same as pointer // to a key location, then we must have found // a key in linked list. We swap data of pCrawl // and pKey and move pKey to next position. if (pCrawl != pKey && pCrawl.data != key) { pKey.data = pCrawl.data; pCrawl.data = key; pKey = pKey.next; } // Find next position where key is present if (pKey.data != key) pKey = pKey.next; // Moving to next Node pCrawl = pCrawl.next; } } // Driver code public static void Main(String[] args) { Node head = newNode(10); head.next = newNode(20); head.next.next = newNode(10); head.next.next.next = newNode(30); head.next.next.next.next = newNode(40); head.next.next.next.next.next = newNode(10); head.next.next.next.next.next.next = newNode(60); Console.Write( "Before moveToEnd(), the Linked list is\n" ); printList(head); int key = 10; moveToEnd(head, key); Console.Write( "\nAfter moveToEnd(), the Linked list is\n" ); printList(head); } } // This code has been contributed by 29AjayKumar |
Output:
Before moveToEnd(), the Linked list is 10 20 10 30 40 10 60 After moveToEnd(), the Linked list is 20 30 40 60 10 10 10
Time Complexity : O(n) requires only one traversal of list.
Efficient Solution 2 :
1. Traverse the linked list and take a pointer at tail.
2. Now, check for the key and node->data, if they are equal, move the node to last-next, else move
ahead.
C++
// C++ code to remove key element to end of linked list #include<bits/stdc++.h> using namespace std; // A Linked list Node struct Node { int data; struct Node* next; }; // A urility function to create a new node. struct Node* newNode( int x) { Node* temp = new Node; temp->data = x; temp->next = NULL; } // Function to remove key to end Node *keyToEnd(Node* head, int key) { // Node to keep pointing to tail Node* tail = head; if (head == NULL) { return NULL; } while (tail->next != NULL) { tail = tail->next; } // Node to point to last of linked list Node* last = tail; Node* current = head; Node* prev = NULL; // Node prev2 to point to previous when head.data!=key Node* prev2 = NULL; // loop to perform operations to remove key to end while (current != tail) { if (current->data == key && prev2 == NULL) { prev = current; current = current->next; head = current; last->next = prev; last = last->next; last->next = NULL; prev = NULL; } else { if (current->data == key && prev2 != NULL) { prev = current; current = current->next; prev2->next = current; last->next = prev; last = last->next; last->next = NULL; } else if (current != tail) { prev2 = current; current = current->next; } } } return head; } // Function to display linked list void printList(Node* head) { struct Node* temp = head; while (temp != NULL) { printf ( "%d " , temp->data); temp = temp->next; } printf ( "\n" ); } // Driver Code int main() { Node* root = newNode(5); root->next = newNode(2); root->next->next = newNode(2); root->next->next->next = newNode(7); root->next->next->next->next = newNode(2); root->next->next->next->next->next = newNode(2); root->next->next->next->next->next->next = newNode(2); int key = 2; cout << "Linked List before operations :" ; printList(root); cout << "\nLinked List after operations :" ; root = keyToEnd(root, key); printList(root); return 0; } // This code is contributed by Rajout-Ji |
Java
// Java code to remove key element to end of linked list import java.util.*; // Node class class Node { int data; Node next; public Node( int data) { this .data = data; this .next = null ; } } class gfg { static Node root; // Function to remove key to end public static Node keyToEnd(Node head, int key) { // Node to keep pointing to tail Node tail = head; if (head == null ) { return null ; } while (tail.next != null ) { tail = tail.next; } // Node to point to last of linked list Node last = tail; Node current = head; Node prev = null ; // Node prev2 to point to previous when head.data!=key Node prev2 = null ; // loop to perform operations to remove key to end while (current != tail) { if (current.data == key && prev2 == null ) { prev = current; current = current.next; head = current; last.next = prev; last = last.next; last.next = null ; prev = null ; } else { if (current.data == key && prev2 != null ) { prev = current; current = current.next; prev2.next = current; last.next = prev; last = last.next; last.next = null ; } else if (current != tail) { prev2 = current; current = current.next; } } } return head; } // Function to display linked list public static void display(Node root) { while (root != null ) { System.out.print(root.data + " " ); root = root.next; } } // Driver Code public static void main(String args[]) { root = new Node( 5 ); root.next = new Node( 2 ); root.next.next = new Node( 2 ); root.next.next.next = new Node( 7 ); root.next.next.next.next = new Node( 2 ); root.next.next.next.next.next = new Node( 2 ); root.next.next.next.next.next.next = new Node( 2 ); int key = 2 ; System.out.println( "Linked List before operations :" ); display(root); System.out.println( "\nLinked List after operations :" ); root = keyToEnd(root, key); display(root); } } |
Python3
# Python3 code to remove key element to # end of linked list # A Linked list Node class Node: def __init__( self , data): self .data = data self . next = None # A utility function to create a new node. def newNode(x): temp = Node(x) return temp # Function to remove key to end def keyToEnd(head, key): # Node to keep pointing to tail tail = head if (head = = None ): return None while (tail. next ! = None ): tail = tail. next # Node to point to last of linked list last = tail current = head prev = None # Node prev2 to point to previous # when head.data!=key prev2 = None # Loop to perform operations to # remove key to end while (current ! = tail): if (current.data = = key and prev2 = = None ): prev = current current = current. next head = current last. next = prev last = last. next last. next = None prev = None else : if (current.data = = key and prev2 ! = None ): prev = current current = current. next prev2. next = current last. next = prev last = last. next last. next = None elif (current ! = tail): prev2 = current current = current. next return head # Function to display linked list def printList(head): temp = head while (temp ! = None ): print (temp.data, end = ' ' ) temp = temp. next print () # Driver Code if __name__ = = '__main__' : root = newNode( 5 ) root. next = newNode( 2 ) root. next . next = newNode( 2 ) root. next . next . next = newNode( 7 ) root. next . next . next . next = newNode( 2 ) root. next . next . next . next . next = newNode( 2 ) root. next . next . next . next . next . next = newNode( 2 ) key = 2 print ( "Linked List before operations :" ) printList(root) print ( "Linked List after operations :" ) root = keyToEnd(root, key) printList(root) # This code is contributed by rutvik_56 |
C#
// C# code to remove key // element to end of linked list using System; // Node class public class Node { public int data; public Node next; public Node( int data) { this .data = data; this .next = null ; } } class GFG { static Node root; // Function to remove key to end public static Node keyToEnd(Node head, int key) { // Node to keep pointing to tail Node tail = head; if (head == null ) { return null ; } while (tail.next != null ) { tail = tail.next; } // Node to point to last of linked list Node last = tail; Node current = head; Node prev = null ; // Node prev2 to point to // previous when head.data!=key Node prev2 = null ; // loop to perform operations // to remove key to end while (current != tail) { if (current.data == key && prev2 == null ) { prev = current; current = current.next; head = current; last.next = prev; last = last.next; last.next = null ; prev = null ; } else { if (current.data == key && prev2 != null ) { prev = current; current = current.next; prev2.next = current; last.next = prev; last = last.next; last.next = null ; } else if (current != tail) { prev2 = current; current = current.next; } } } return head; } // Function to display linked list public static void display(Node root) { while (root != null ) { Console.Write(root.data + " " ); root = root.next; } } // Driver Code public static void Main() { root = new Node(5); root.next = new Node(2); root.next.next = new Node(2); root.next.next.next = new Node(7); root.next.next.next.next = new Node(2); root.next.next.next.next.next = new Node(2); root.next.next.next.next.next.next = new Node(2); int key = 2; Console.WriteLine( "Linked List before operations :" ); display(root); Console.WriteLine( "\nLinked List after operations :" ); root = keyToEnd(root, key); display(root); } } // This code is contributed by PrinciRaj1992 |
Output:
Linked List before operations : 5 2 2 7 2 2 2 Linked List after operations : 5 7 2 2 2 2 2
Thanks to Ravinder Kumar for suggesting this method.
Efficient Solution 3 : is to maintain a separate list of keys. We initialize this list of keys as empty. We traverse given list. For every key found, we remove it from the original list and insert into the separate list of keys. We finally link list of keys at the end of remaining given list. Time complexity of this solution is also O(n) and it also requires only one traversal of list.
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