[LeetCode] Swap Nodes in Pairs

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.

Solution: The power of recursion
My first solution is not recursive, yet it is not only bug-prone, but also hard to follow.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode p = head;
        head = head.next;
        ListNode tail = null;
        while (p != null && p.next != null) {
            ListNode q = p.next;
            ListNode r = q.next;
            q.next = p;
            p.next = r;
            if (tail != null) {
                tail.next = q;
            }
            tail = p;
            p = r;
        }
        return head;
    }
}

My second solution is recursive, and it is much easier to follow and to implement!

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode p = head, q = head.next, r = q.next;
        q.next = p;
        p.next = swapPairs(r);
        return q;
    }
}