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->14
This may look 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.
Python
# 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 a linked list by changing links def swapNodes( self , x, y): # Nothing to do if x and y are # the 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 code 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 " Linked list after calling swapNodes() " llist.printList() # This code is contributed by BHAVYA JAIN |
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
Time Complexity: O(n)
Auxiliary Space: O(1)
Optimizations: The above code can be optimized to search x and y in single traversal. Two loops are used to keep program simple.
Simpler approach:
Python
# 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 their 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 |
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
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Swap nodes in a linked list without swapping data for more details!
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