Leetcode #1721: Swapping Nodes in a Linked List
In this guide, we solve Leetcode #1721 Swapping Nodes in a Linked List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Linked List, Two Pointers
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: head = [1,2,3,4,5], k = 2
Output: [1,4,3,2,5]
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
fast = slow = head
for _ in range(k - 1):
fast = fast.next
p = fast
while fast.next:
fast, slow = fast.next, slow.next
q = slow
p.val, q.val = q.val, p.val
return head
Complexity
The time complexity is , where is the length of the linked list. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.