Swap Every Kth Node in a LinkedList
Objective: Given a linked list, swap every kth node in that. If at the end of the list remaining nodes are less than k, leave them untouched.
Input: A linked list, A number k.
Example:
Input : ->1->2->3->4->5->6->7->8->9->10 , K = 4 Output: ->4->2->3->1->8->6->7->5->9->10
Approach:
- Take 3 Pointers, ptrOne, ptrTwo and ptrTwo_prev.
- ptrOne and ptrTwo_prev points at head node.
- ptrTwo points at next node of ptrTwo_prev.
- Move the ptrTwo and ptrTwo_prev k-2 times, since we need one pointer each at both ends for swapping so move pointers only k-2 times.
- Create another pointer , NewHead and point it to ptrTwo.next.
- Now we have ptrOne at head and ptrTwo at kth position, swap them with the help of ptrTwo_prev.
- This function will returns the head.
- Now make a recursive call with newHead.
ptrOne.next = reverseNodes(newHead, k);
Swap Evey Kth Node in a linked list
Complete Code:
Output: Original Link List 1 : ->1->2->3->4->5->6->7->8->9->10 Swap Every 4th Node : ->4->2->3->1->8->6->7->5->9->10
There is one more little easy solution of the problem-:
private void swap(Node head,int k) {
if(head==null || k<2){
return;
}
int i = 2;
Node first = head;
Node second = null;
Node curr = head.next;
while (curr != null) {
if (i % k == 0) {
second = curr;
int temp = first.data;
first.data = second.data;
second.data = temp;
first = second.next;
}
curr=curr.next;
i++;
}
}