14. Swapping Nodes in a Linked List.cpp

/*
Swapping Nodes in a Linked List
===============================

You are given the head of a linked list, and an integer k.
Return the head of the linked list after swapping the values of the kth node from the beginning and the kth node from the end (the list is 1-indexed).

Example 1:
Input: head = [1,2,3,4,5], k = 2
Output: [1,4,3,2,5]

Example 2:
Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
Output: [7,9,6,6,8,7,3,0,9,5]

Example 3:
Input: head = [1], k = 1
Output: [1]

Example 4:
Input: head = [1,2], k = 1
Output: [2,1]

Example 5:
Input: head = [1,2,3], k = 2
Output: [1,2,3]

Constraints:
The number of nodes in the list is n.
1 <= k <= n <= 105
0 <= Node.val <= 100

Hint #1  
We can transform the linked list to an array this should ease things up

Hint #2  
After transforming the linked list to an array it becomes as easy as swapping two integers in an array then rebuilding the linked list
*/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution
{
public:
  ListNode *swapNodes(ListNode *head, int k)
  {
    int size = 0;
    auto temp = head;
    while (temp)
    {
      temp = temp->next;
      size++;
    }

    auto front = head, back = head;
    for (int i = 0; i < k - 1; ++i)
      front = front->next;
    for (int i = 0; i < size - k; ++i)
      back = back->next;
    swap(front->val, back->val);
    return head;
  }
};