Modify contents of Linked List
Given a singly linked list containing n nodes. Modify the value of first half nodes such that 1st node’s new value is equal to the last node’s value minus first node’s current value, 2nd node’s new value is equal to the second last node’s value minus 2nd node’s current value, likewise for first half nodes. If n is odd then the value of the middle node remains unchanged.
[No extra memory to be used].
Examples:
Input : 10 -> 4 -> 5 -> 3 -> 6 Output : 4 -> 1 -> 5 -> 3 -> 6 Input : 2 -> 9 -> 8 -> 12 -> 7 -> 10 Output : -8 -> 2 -> -4 -> 12 -> 7 -> 10Asked in Amazon Interview
Swap nodes in a linked list without swapping data
Given a linked list and two keys in it, swap nodes for two given keys. Nodes should be swapped by changing links. Swapping data of nodes may be expensive in many situations when data contains many fields.
It may be assumed that all keys in the linked list are distinct.
Examples:
Input : 10->15->12->13->20->14, x = 12, y = 20 Output: 10->15->20->13->12->14 Input : 10->15->12->13->20->14, x = 10, y = 20 Output: 20->15->12->13->10->14 Input : 10->15->12->13->20->14, x = 12, y = 13 Output: 10->15->13->12->20->14This may look like a simple problem, but is an interesting question as it has the following cases to be handled.
- x and y may or may not be adjacent.
- Either x or y may be a head node.
- Either x or y may be the last node.
- x and/or y may not be present in the linked list.
How to write a clean working code that handles all the above possibilities.
The idea is to first search x and y in the given linked list. If any of them is not present, then return. While searching for x and y, keep track of current and previous pointers. First change next of previous pointers, then change next of current pointers.
Below is the implementation of the above approach.
C++
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/* This program swaps the nodes of linked list rather than swapping the field from the nodes.*/ #include using namespace std; /* A linked list node */ class Node { public: int data; Node* next; }; /* Function to swap nodes x and y in linked list by changing links */ void swapNodes[Node** head_ref, int x, int y] { // Nothing to do if x and y are same if [x == y] return; // Search for x [keep track of prevX and CurrX Node *prevX = NULL, *currX = *head_ref; while [currX && currX->data != x] { prevX = currX; currX = currX->next; } // Search for y [keep track of prevY and CurrY Node *prevY = NULL, *currY = *head_ref; while [currY && currY->data != y] { prevY = currY; currY = currY->next; } // If either x or y is not present, nothing to do if [currX == NULL || currY == NULL] return; // If x is not head of linked list if [prevX != NULL] prevX->next = currY; else // Else make y as new head *head_ref = currY; // If y is not head of linked list if [prevY != NULL] prevY->next = currX; else // Else make x as new head *head_ref = currX; // Swap next pointers Node* temp = currY->next; currY->next = currX->next; currX->next = temp; } /* Function to add a node at the beginning of 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 nodes in a given linked list */ void printList[Node* node] { while [node != NULL] { cout data next; } } /* Driver program to test above function */ int main[] { Node* start = NULL; /* The constructed linked list is: 1->2->3->4->5->6->7 */ push[&start, 7]; push[&start, 6]; push[&start, 5]; push[&start, 4]; push[&start, 3]; push[&start, 2]; push[&start, 1]; cout next; } // Search for y [keep track of prevY and CurrY struct Node *prevY = NULL, *currY = *head_ref; while [currY && currY->data != y] { prevY = currY; currY = currY->next; } // If either x or y is not present, nothing to do if [currX == NULL || currY == NULL] return; // If x is not head of linked list if [prevX != NULL] prevX->next = currY; else // Else make y as new head *head_ref = currY; // If y is not head of linked list if [prevY != NULL] prevY->next = currX; else // Else make x as new head *head_ref = currX; // Swap next pointers struct Node* temp = currY->next; currY->next = currX->next; currX->next = temp; } /* Function to add a node at the beginning of 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 function */ int main[] { struct Node* start = NULL; /* The constructed linked list is: 1->2->3->4->5->6->7 */ push[&start, 7]; push[&start, 6]; push[&start, 5]; push[&start, 4]; push[&start, 3]; push[&start, 2]; push[&start, 1]; printf["\n Linked list before calling swapNodes[] "]; printList[start]; swapNodes[&start, 4, 3]; printf["\n Linked list after calling swapNodes[] "]; printList[start]; return 0; } |
Java
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// Java program to swap two given nodes of a linked list class Node { int data; Node next; Node[int d] { data = d; next = null; } } class LinkedList { Node head; // head of list /* Function to swap Nodes x and y in linked list by changing links */ public void swapNodes[int x, int y] { // Nothing to do if x and y are same if [x == y] return; // Search for x [keep track of prevX and CurrX] Node prevX = null, currX = head; while [currX != null && currX.data != x] { prevX = currX; currX = currX.next; } // Search for y [keep track of prevY and currY] Node prevY = null, currY = head; while [currY != null && currY.data != y] { prevY = currY; currY = currY.next; } // If either x or y is not present, nothing to do if [currX == null || currY == null] return; // If x is not head of linked list if [prevX != null] prevX.next = currY; else // make y the new head head = currY; // If y is not head of linked list if [prevY != null] prevY.next = currX; else // make x the new head head = currX; // Swap next pointers Node temp = currX.next; currX.next = currY.next; currY.next = temp; } /* Function to add Node at beginning of list. */ public void push[int new_data] { /* 1. alloc the Node and put the data */ Node new_Node = new Node[new_data]; /* 2. Make next of new Node as head */ new_Node.next = head; /* 3. Move the head to point to new Node */ head = new_Node; } /* This function prints contents of linked list starting from the given Node */ public void printList[] { Node tNode = head; while [tNode != null] { System.out.print[tNode.data + " "]; tNode = tNode.next; } } /* Driver program to test above function */ public static void main[String[] args] { LinkedList llist = new LinkedList[]; /* The constructed linked list is: 1->2->3->4->5->6->7 */ llist.push[7]; llist.push[6]; llist.push[5]; llist.push[4]; llist.push[3]; llist.push[2]; llist.push[1]; System.out.print[ "\n Linked list before calling swapNodes[] "]; llist.printList[]; llist.swapNodes[4, 3]; System.out.print[ "\n Linked list after calling swapNodes[] "]; llist.printList[]; } } // This code is contributed by Rajat Mishra |
Python
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# Python program to swap two given nodes of a linked list class LinkedList[object]: def __init__[self]: self.head = None # head of list class Node[object]: def __init__[self, d]: self.data = d self.next = None # Function to swap Nodes x and y in linked list by # changing links def swapNodes[self, x, y]: # Nothing to do if x and y are same if x == y: return # Search for x [keep track of prevX and CurrX] prevX = None currX = self.head while currX != None and currX.data != x: prevX = currX currX = currX.next # Search for y [keep track of prevY and currY] prevY = None currY = self.head while currY != None and currY.data != y: prevY = currY currY = currY.next # If either x or y is not present, nothing to do if currX == None or currY == None: return # If x is not head of linked list if prevX != None: prevX.next = currY else: # make y the new head self.head = currY # If y is not head of linked list if prevY != None: prevY.next = currX else: # make x the new head self.head = currX # Swap next pointers temp = currX.next currX.next = currY.next currY.next = temp # Function to add Node at beginning of list. def push[self, new_data]: # 1. alloc the Node and put the data new_Node = self.Node[new_data] # 2. Make next of new Node as head new_Node.next = self.head # 3. Move the head to point to new Node self.head = new_Node # This function prints contents of linked list starting # from the given Node def printList[self]: tNode = self.head while tNode != None: print tNode.data, tNode = tNode.next # Driver program to test above function llist = LinkedList[] # The constructed linked list is: # 1->2->3->4->5->6->7 llist.push[7] llist.push[6] llist.push[5] llist.push[4] llist.push[3] llist.push[2] llist.push[1] print "Linked list before calling swapNodes[] " llist.printList[] llist.swapNodes[4, 3] print "\nLinked list after calling swapNodes[] " llist.printList[] # This code is contributed by BHAVYA JAIN |
C#
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// C# program to swap two given // nodes of a linked list using System; class Node { public int data; public Node next; public Node[int d] { data = d; next = null; } } public class LinkedList { Node head; // head of list /* Function to swap Nodes x and y in linked list by changing links */ public void swapNodes[int x, int y] { // Nothing to do if x and y are same if [x == y] return; // Search for x [keep track of prevX and CurrX] Node prevX = null, currX = head; while [currX != null && currX.data != x] { prevX = currX; currX = currX.next; } // Search for y [keep track of prevY and currY] Node prevY = null, currY = head; while [currY != null && currY.data != y] { prevY = currY; currY = currY.next; } // If either x or y is not present, nothing to do if [currX == null || currY == null] return; // If x is not head of linked list if [prevX != null] prevX.next = currY; else // make y the new head head = currY; // If y is not head of linked list if [prevY != null] prevY.next = currX; else // make x the new head head = currX; // Swap next pointers Node temp = currX.next; currX.next = currY.next; currY.next = temp; } /* Function to add Node at beginning of list. */ public void push[int new_data] { /* 1. alloc the Node and put the data */ Node new_Node = new Node[new_data]; /* 2. Make next of new Node as head */ new_Node.next = head; /* 3. Move the head to point to new Node */ head = new_Node; } /* This function prints contents of linked list starting from the given Node */ public void printList[] { Node tNode = head; while [tNode != null] { Console.Write[tNode.data + " "]; tNode = tNode.next; } } /* Driver code */ public static void Main[String[] args] { LinkedList llist = new LinkedList[]; /* The constructed linked list is: 1->2->3->4->5->6->7 */ llist.push[7]; llist.push[6]; llist.push[5]; llist.push[4]; llist.push[3]; llist.push[2]; llist.push[1]; Console.Write[ "\n Linked list before calling swapNodes[] "]; llist.printList[]; llist.swapNodes[4, 3]; Console.Write[ "\n Linked list after calling swapNodes[] "]; llist.printList[]; } } // This code is contributed by 29AjayKumar |
Javascript
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// JavaScript program to swap two // given nodes of a linked list class Node { constructor[val] { this.data = val; this.next = null; } } var head; // head of list /* Function to swap Nodes x and y in linked list by changing links */ function swapNodes[x , y] { // Nothing to do if x and y are same if [x == y] return; // Search for x [keep track of prevX and CurrX] var prevX = null, currX = head; while [currX != null && currX.data != x] { prevX = currX; currX = currX.next; } // Search for y [keep track of prevY and currY] var prevY = null, currY = head; while [currY != null && currY.data != y] { prevY = currY; currY = currY.next; } // If either x or y is not present, nothing to do if [currX == null || currY == null] return; // If x is not head of linked list if [prevX != null] prevX.next = currY; else // make y the new head head = currY; // If y is not head of linked list if [prevY != null] prevY.next = currX; else // make x the new head head = currX; // Swap next pointers var temp = currX.next; currX.next = currY.next; currY.next = temp; } /* Function to add Node at beginning of list. */ function push[new_data] { /* 1. alloc the Node and put the data */ var new_Node = new Node[new_data]; /* 2. Make next of new Node as head */ new_Node.next = head; /* 3. Move the head to point to new Node */ head = new_Node; } /* This function prints contents of linked list starting from the given Node */ function printList[] { var tNode = head; while [tNode != null] { document.write[tNode.data + " "]; tNode = tNode.next; } } /* Driver program to test above function */
/* The constructed linked list is: 1->2->3->4->5->6->7 */ push[7]; push[6]; push[5]; push[4]; push[3]; push[2]; push[1]; document.write[
" Linked list before calling swapNodes[] ; printList[]; swapNodes[4, 3]; document.write[
" ]; printList[]; // This code is contributed by todaysgaurav |
Output Linked list before calling swapNodes[] 1 2 3 4 5 6 7 Linked list after calling swapNodes[] 1 2 4 3 5 6 7
Optimizations: The above code can be optimized to search x and y in a single traversal. Two loops are used to keep the program simple.
Simpler approach –
C++
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// C++ program to swap two given nodes of a linked list #include using namespace std; // A linked list node class class Node { public: int data; class Node* next; // constructor Node[int val, Node* next] : data[val] , next[next] { } // print list from this // to last till null void printList[] { Node* node = this; while [node != NULL] { cout data next; } cout data == x] { a = head_ref; } else if [[*head_ref]->data == y] { b = head_ref; } head_ref = &[[*head_ref]->next]; } // if we have found both a and b // in the linked list swap current // pointer and next pointer of these if [a && b] { swap[*a, *b]; swap[[[*a]->next], [[*b]->next]]; } } // Driver code int main[] { Node* start = NULL; // The constructed linked list is: // 1->2->3->4->5->6->7 push[&start, 7]; push[&start, 6]; push[&start, 5]; push[&start, 4]; push[&start, 3]; push[&start, 2]; push[&start, 1]; cout printList[]; swapNodes[&start, 6, 1]; cout printList[]; } |
Java
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// Java program to swap two given nodes of a linked list public class Solution { // Represent a node of the singly linked list class Node { int data; Node next; public Node[int data] { this.data = data; this.next = null; } } // Represent the head and tail of the singly linked list public Node head = null; public Node tail = null; // addNode[] will add a new node to the list public void addNode[int data] { // Create a new node Node newNode = new Node[data]; // Checks if the list is empty if [head == null] { // If list is empty, both head and // tail will point to new node head = newNode; tail = newNode; } else { // newNode will be added after tail such that // tail's next will point to newNode tail.next = newNode; // newNode will become new tail of the list tail = newNode; } } // swap[] will swap the given two nodes public void swap[int n1, int n2] { Node prevNode1 = null, prevNode2 = null, node1 = head, node2 = head; // Checks if list is empty if [head == null] { return; } // If n1 and n2 are equal, then // list will remain the same if [n1 == n2] return; // Search for node1 while [node1 != null && node1.data != n1] { prevNode1 = node1; node1 = node1.next; } // Search for node2 while [node2 != null && node2.data != n2] { prevNode2 = node2; node2 = node2.next; } if [node1 != null && node2 != null] { // If previous node to node1 is not null then, // it will point to node2 if [prevNode1 != null] prevNode1.next = node2; else head = node2; // If previous node to node2 is not null then, // it will point to node1 if [prevNode2 != null] prevNode2.next = node1; else head = node1; // Swaps the next nodes of node1 and node2 Node temp = node1.next; node1.next = node2.next; node2.next = temp; } else { System.out.println["Swapping is not possible"]; } } // display[] will display all the // nodes present in the list public void display[] { // Node current will point to head Node current = head; if [head == null] { System.out.println["List is empty"]; return; } while [current != null] { // Prints each node by incrementing pointer System.out.print[current.data + " "]; current = current.next; } System.out.println[]; } public static void main[String[] args] { Solution sList = new Solution[]; // Add nodes to the list sList.addNode[1]; sList.addNode[2]; sList.addNode[3]; sList.addNode[4]; sList.addNode[5]; sList.addNode[6]; sList.addNode[7]; System.out.println["Original list: "]; sList.display[]; // Swaps the node 2 with node 5 sList.swap[6, 1]; System.out.println["List after swapping nodes: "]; sList.display[]; } } |
Python3
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# Python3 program to swap two given # nodes of a linked list # A linked list node class class Node: # constructor def __init__[self, val=None, next1=None]: self.data = val self.next = next1 # print list from this # to last till None def printList[self]: node = self while [node != None]: print[node.data, end=" "] node = node.next print[" "] # Function to add a node # at the beginning of List def push[head_ref, new_data]: # allocate node [head_ref] = Node[new_data, head_ref] return head_ref def swapNodes[head_ref, x, y]: head = head_ref # Nothing to do if x and y are same if [x == y]: return None a = None b = None # search for x and y in the linked list # and store therir pointer in a and b while [head_ref.next != None]: if [[head_ref.next].data == x]: a = head_ref elif [[head_ref.next].data == y]: b = head_ref head_ref = [[head_ref].next] # if we have found both a and b # in the linked list swap current # pointer and next pointer of these if [a != None and b != None]: temp = a.next a.next = b.next b.next = temp temp = a.next.next a.next.next = b.next.next b.next.next = temp return head # Driver code start = None # The constructed linked list is: # 1.2.3.4.5.6.7 start = push[start, 7] start = push[start, 6] start = push[start, 5] start = push[start, 4] start = push[start, 3] start = push[start, 2] start = push[start, 1] print["Linked list before calling swapNodes[] "] start.printList[] start = swapNodes[start, 6, 1] print["Linked list after calling swapNodes[] "] start.printList[] # This code is contributed by Arnab Kundu |
C#
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// C# program to swap two // given nodes of a linked list using System; class GFG { // A linked list node class public class Node { public int data; public Node next; // constructor public Node[int val, Node next1] { data = val; next = next1; } // print list from this // to last till null public void printList[] { Node node = this; while [node != null] { Console.Write[node.data + " "]; node = node.next; } Console.WriteLine[]; } } // Function to add a node // at the beginning of List static Node push[Node head_ref, int new_data] { // allocate node [head_ref] = new Node[new_data, head_ref]; return head_ref; } static Node swapNodes[Node head_ref, int x, int y] { Node head = head_ref; // Nothing to do if x and y are same if [x == y] return null; Node a = null, b = null; // search for x and y in the linked list // and store therir pointer in a and b while [head_ref.next != null] { if [[head_ref.next].data == x] { a = head_ref; } else if [[head_ref.next].data == y] { b = head_ref; } head_ref = [[head_ref].next]; } // if we have found both a and b // in the linked list swap current // pointer and next pointer of these if [a != null && b != null] { Node temp = a.next; a.next = b.next; b.next = temp; temp = a.next.next; a.next.next = b.next.next; b.next.next = temp; } return head; } // Driver code public static void Main[] { Node start = null; // The constructed linked list is: // 1.2.3.4.5.6.7 start = push[start, 7]; start = push[start, 6]; start = push[start, 5]; start = push[start, 4]; start = push[start, 3]; start = push[start, 2]; start = push[start, 1]; Console.Write[ "Linked list before calling swapNodes[] "]; start.printList[]; start = swapNodes[start, 6, 1]; Console.Write[ "Linked list after calling swapNodes[] "]; start.printList[]; } } /* This code contributed by PrinciRaj1992 */ |
Javascript
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// javascript program to swap two given nodes of a linked list // Represent a node of the singly linked list class Node { constructor[val] { this.data = val; this.next = null; } }
// Represent the head and tail of the singly linked list var head = null; var tail = null; // addNode[] will add a new node to the list function addNode[data] { // Create a new node var newNode = new Node[data]; // Checks if the list is empty if [head == null] { // If list is empty, both head and // tail will point to new node head = newNode; tail = newNode; } else { // newNode will be added after tail such that // tail's next will point to newNode tail.next = newNode; // newNode will become new tail of the list tail = newNode; } } // swap[] will swap the given two nodes function swap[n1 , n2] { var prevNode1 = null, prevNode2 = null, node1 = head, node2 = head; // Checks if list is empty if [head == null] { return; } // If n1 and n2 are equal, then // list will remain the same if [n1 == n2] return; // Search for node1 while [node1 != null && node1.data != n1] { prevNode1 = node1; node1 = node1.next; } // Search for node2 while [node2 != null && node2.data != n2] { prevNode2 = node2; node2 = node2.next; } if [node1 != null && node2 != null] { // If previous node to node1 is not null then, // it will point to node2 if [prevNode1 != null] prevNode1.next = node2; else head = node2; // If previous node to node2 is not null then, // it will point to node1 if [prevNode2 != null] prevNode2.next = node1; else head = node1; // Swaps the next nodes of node1 and node2 var temp = node1.next; node1.next = node2.next; node2.next = temp; } else { document.write["Swapping is not possible"]; } } // display[] will display all the // nodes present in the list function display[] { // Node current will point to head var current = head; if [head == null] { document.write["List is empty"]; return; } while [current != null] { // Prints each node by incrementing pointer document.write[current.data + " "]; current = current.next; } document.write[]; }
// Add nodes to the list addNode[1]; addNode[2]; addNode[3]; addNode[4]; addNode[5]; addNode[6]; addNode[7];
document.write["Original list: display[]; // Swaps the node 2 with node 5 swap[6, 1];
document.write[" display[]; // This code contributed by aashish2995 |
Output Linked list before calling swapNodes[] 1 2 3 4 5 6 7 Linked list after calling swapNodes[] 6 2 3 4 5 1 7
//www.youtube.com/watch?v=V4ZHvhvVmSE
This article is contributed by Gautam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Article Tags :
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 | #include #include // A Linked List Node struct Node { int data; struct Node* next; }; // Helper function to print a given linked list void printList[struct Node* head] { struct Node* ptr = head; while [ptr] { printf["%d —> ", ptr->data]; ptr = ptr->next; } printf["NULL\n"]; } // Helper function to insert a new node at the beginning of the linked list void push[struct Node** head, int data] { struct Node* newNode = [struct Node*]malloc[sizeof[struct Node]]; newNode->data = data; newNode->next = *head; *head = newNode; } // Function takes the node from the front of the source and move it // to the front of the destination void moveNode[struct Node** destRef, struct Node** sourceRef] { // if the source list empty, do nothing if [*sourceRef == NULL] { return; } struct Node* newNode = *sourceRef;// the front source node *sourceRef = [*sourceRef]->next;// advance the source pointer newNode->next = *destRef; // link the old dest off the new node *destRef = newNode; // move dest to point to the new node } int main[void] { // input keys int keys[] = { 1, 2, 3 }; int n = sizeof[keys]/sizeof[keys[0]]; // construct the first linked list struct Node* a = NULL; for [int i = n-1; i >= 0; i--] { push[&a, keys[i]]; } // construct the second linked list struct Node* b = NULL; for [int i = 0; i 1 —> 2 —> 3 —> NULL C
DownloadRun Code Output: Linked List Operations: Traverse, Insert and DeleteIn this tutorial, you will learn different operations on a linked list. Also, you will find implementation of linked list operations in C/C++, Python and Java. There are various linked list operations that allow us to perform different actions on linked lists. For example, the insertion operation adds a new element to the linked list. Here's a list of basic linked list operations that we will cover in this article.
Before you learn about linked list operations in detail, make sure to know about Linked List first. Things to Remember about Linked List
In all of the examples, we will assume that the linked list has three nodes 1 --->2 --->3 with node structure as below: struct node { int data; struct node *next; };C++
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