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 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 interesting question as it has following cases to be handled.
1) x and y may or may not be adjacent.
2) Either x or y may be a head node.
3) Either x or y may be last node.
4) x and/or y may not be present in linked list.

[ad type=”banner”]

How to write a clean working code that handles all of the above possibilities.

The idea it to first search x and y in 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. Following python implementation of this approach.

Python Programming:

# 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

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 single traversal. Two loops are used to keep program simple.

[ad type=”banner”]

Categorized in:

Linked ListPYTHONSingly Linked List

Tagged in:

, , , , , , , , , , ,