Swap Nodes in Pairs LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Swap Nodes in Pairs in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Swap Nodes in Pairs– LeetCode Problem

Swap Nodes in Pairs– LeetCode Problem

Problem:

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list’s nodes (i.e., only nodes themselves may be changed.)

Example 1:

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

Example 2:

Input: head = []
Output: []

Example 3:

Input: head = [1]
Output: [1]

Constraints:

• The number of nodes in the list is in the range [0, 100].
• 0 <= Node.val <= 100

Swap Nodes in Pairs– LeetCode Solutions

class Solution {
 public:
  ListNode* swapPairs(ListNode* head) {
    const int length = getLength(head);
    ListNode dummy(0, head);
    ListNode* prev = &dummy;
    ListNode* curr = head;

    for (int i = 0; i < length / 2; ++i) {
      ListNode* next = curr->next;
      curr->next = next->next;
      next->next = prev->next;
      prev->next = next;
      prev = curr;
      curr = curr->next;
    }

    return dummy.next;
  }

 private:
  int getLength(ListNode* head) {
    int length = 0;
    for (ListNode* curr = head; curr; curr = curr->next)
      ++length;
    return length;
  }
};

class Solution {
  public ListNode swapPairs(ListNode head) {
    final int length = getLength(head);
    ListNode dummy = new ListNode(0, head);
    ListNode prev = dummy;
    ListNode curr = head;

    for (int i = 0; i < length / 2; ++i) {
      ListNode next = curr.next;
      curr.next = next.next;
      next.next = curr;
      prev.next = next;
      prev = curr;
      curr = curr.next;
    }

    return dummy.next;
  }

  private int getLength(ListNode head) {
    int length = 0;
    for (ListNode curr = head; curr != null; curr = curr.next)
      ++length;
    return length;
  }
}

class Solution:
  def swapPairs(self, head: ListNode) -> ListNode:
    def getLength(head: ListNode) -> int:
      length = 0
      while head:
        length += 1
        head = head.next
      return length

    length = getLength(head)
    dummy = ListNode(0, head)
    prev = dummy
    curr = head

    for _ in range(length // 2):
      next = curr.next
      curr.next = next.next
      next.next = prev.next
      prev.next = next
      prev = curr
      curr = curr.next

    return dummy.next