24. Swap Nodes in Pairs

Summary

Solution Approach	Explanation (1-minute)	Time Complexity	Space Complexity
Iterative Pointer Manipulation	Use a dummy head node and maintain three pointers (prev, curr, next) to iteratively swap adjacent pairs. For each pair, redirect pointers to swap positions. Continue until no more pairs exist.	\(O(n)\)	\(O(1)\)

Example

   [0 -> 1 -> 2 -> 3 -> 4]
-> [0 -> 2 -> 1 -> 3 -> 4]
-> [0 -> 2 -> 1 -> 4 -> 3]

Implementation

PythonJavaScriptGo

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        preHead = ListNode(None, head)

        prev = preHead
        curr = head

        while curr and curr.next:
            nxt = curr.next
            aftNxt = curr.next.next

            prev.next = nxt
            curr.next = aftNxt
            nxt.next = curr

            prev = curr
            curr = aftNxt

        return preHead.next