LeetCode-in-All

24. Swap Nodes in Pairs

Medium

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:

To solve the “Swap Nodes in Pairs” problem in Swift with a Solution class, we can traverse the linked list while swapping pairs of nodes. Here are the steps:

  1. Define a Solution class.
  2. Define a method named swapPairs that takes the head of a linked list as input and returns the head of the modified list.
  3. Create a dummy ListNode object and set its next pointer to the head of the input list. This dummy node will serve as the new head of the modified list.
  4. Initialize three pointers: prev, first, and second.
  5. Iterate through the list while first and second are not null:
    • Assign first to the next pointer of prev.
    • Assign second to the next pointer of first.
    • Assign the next pointer of prev to the next pointer of second.
    • Assign the next pointer of second to first.
    • Move prev to first.
    • Move first to first.next (which is the next pair of nodes).
  6. Return the next pointer of the dummy node, which points to the head of the modified list.

Here’s the implementation:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
public class Solution {
    public func swapPairs(_ head: ListNode?) -> ListNode? {
        if head == nil {
            return nil
        }
        let len = getLength(head)
        return reverse(head, len)
    }

    private func getLength(_ curr: ListNode?) -> Int {
        var cnt = 0
        var curr = curr
        while curr != nil {
            cnt += 1
            curr = curr?.next
        }
        return cnt
    }

    private func reverse(_ head: ListNode?, _ len: Int) -> ListNode? {
        if len < 2 {
            return head
        }
        var curr = head
        var prev: ListNode? = nil
        var next: ListNode?
        for _ in 0..<2 {
            next = curr?.next
            curr?.next = prev
            prev = curr
            curr = next
        }
        head?.next = reverse(curr, len - 2)
        return prev
    }
}

This implementation provides a solution to the “Swap Nodes in Pairs” problem in Swift without modifying the values in the list’s nodes.