**Swap Nodes in Pairs LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Problems C++**

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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

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