24. Swap Nodes in Pairs

Problem description:

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

You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example:

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

Solution:

Create a dummy node and use slow, fast to do swap.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null) return head;
        ListNode fast, slow, dummy;
        slow = new ListNode(); slow.next = head; dummy = slow;
        fast = head;
        
        while(fast != null && fast.next != null){
            slow.next = fast.next;
            fast.next = slow.next.next;
            slow.next.next = fast;
            fast = fast.next;
            slow = slow.next.next;
        }
        return dummy.next;
    }
}

time complexity: $O(n)$
space complexity: $O(1)$
reference:

related problem:
25. Reverse Nodes in k-Group