C++ Program For Swapping Kth Node From Beginning With Kth Node From End In A Linked List
Given a singly linked list, swap kth node from beginning with kth node from end. Swapping of data is not allowed, only pointers should be changed. This requirement may be logical in many situations where the linked list data part is huge (For example student details line Name, RollNo, Address, ..etc). The pointers are always fixed (4 bytes for most of the compilers).
Example:
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 2
Output: 1 -> 4 -> 3 -> 2 -> 5
Explanation: The 2nd node from 1st is 2 and
2nd node from last is 4, so swap them.
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 5
Output: 5 -> 2 -> 3 -> 4 -> 1
Explanation: The 5th node from 1st is 5 and
5th node from last is 1, so swap them.
Illustration:
Approach: The idea is very simple find the k th node from the start and the kth node from last is n-k+1 th node from start. Swap both the nodes.Â
However there are some corner cases, which must be handled
- Y is next to X
- X is next to Y
- X and Y are same
- X and Y don’t exist (k is more than number of nodes in linked list)
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node* next;
};
void push( struct Node** head_ref,
int new_data)
{
struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct Node* node)
{
while (node != NULL)
{
cout << node->data << " " ;
node = node->next;
}
cout << endl;
}
int countNodes( struct Node* s)
{
int count = 0;
while (s != NULL)
{
count++;
s = s->next;
}
return count;
}
void swapKth( struct Node** head_ref, int k)
{
int n = countNodes(*head_ref);
if (n < k)
return ;
if (2 * k - 1 == n)
return ;
Node* x = *head_ref;
Node* x_prev = NULL;
for ( int i = 1; i < k; i++)
{
x_prev = x;
x = x->next;
}
Node* y = *head_ref;
Node* y_prev = NULL;
for ( int i = 1; i < n - k + 1; i++)
{
y_prev = y;
y = y->next;
}
if (x_prev)
x_prev->next = y;
if (y_prev)
y_prev->next = x;
Node* temp = x->next;
x->next = y->next;
y->next = temp;
if (k == 1)
*head_ref = y;
if (k == n)
*head_ref = x;
}
int main()
{
struct Node* head = NULL;
for ( int i = 8; i >= 1; i--)
push(&head, i);
cout << "Original Linked List: " ;
printList(head);
for ( int k = 1; k < 9; k++)
{
swapKth(&head, k);
cout <<
"Modified List for k = " << k << endl;
printList(head);
}
return 0;
}
|
Output:
Original Linked List: 1 2 3 4 5 6 7 8
Modified List for k = 1
8 2 3 4 5 6 7 1
Modified List for k = 2
8 7 3 4 5 6 2 1
Modified List for k = 3
8 7 6 4 5 3 2 1
Modified List for k = 4
8 7 6 5 4 3 2 1
Modified List for k = 5
8 7 6 4 5 3 2 1
Modified List for k = 6
8 7 3 4 5 6 2 1
Modified List for k = 7
8 2 3 4 5 6 7 1
Modified List for k = 8
1 2 3 4 5 6 7 8
Complexity Analysis:
- Time Complexity: O(n), where n is the length of the list.Â
One traversal of the list is needed.
- Auxiliary Space: O(1).Â
No extra space is required.
Please refer complete article on Swap Kth node from beginning with Kth node from end in a Linked List for more details!
Last Updated :
23 Dec, 2021
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