Leetcode 24. Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

Analysis:

Have a dummy node and a temporal node. Iterate the whole list and change every node pair.

Nice to have the figure to help understand.

Time complexity: O(n)

Space complexity: O(1)

Code is below:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        // O(n)
        
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        ListNode node = dummy;
        dummy.next = head;
        
        while (head != null && head.next != null) {
            // First operation
            node.next = head.next;
            head.next = node.next.next;
            node.next.next = head;
            
            // Go to another round
            node = head;
            head = head.next;
        }
        
        return dummy.next;
    }
}