LeetCode Swap Nodes in Pairs

原题链接在这里：https://leetcode.com/problems/swap-nodes-in-pairs/

题目：

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. 

题解:

需要建一个dunmy，最后return dunmy.next. 如此可以避免边界问题的讨论，例如头两个点的转换。

Time Complexity: O(n), n是list的长度. Space: O(1).

AC Java:

/**
 * 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) {
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode mark = dummy;
        ListNode A;
        ListNode B;
        while(mark.next != null && mark.next.next != null){
            A = mark.next;
            B = mark.next.next;
            A.next = B.next;
            B.next = A;
            mark.next = B;
            mark = mark.next.next;
        }
        return dummy.next;
    }
}

Recursion解法.

Time Complexity: O(n). Space: O(n), stack space.

AC Java: 

/**
 * 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) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode n = head.next;
        head.next = swapPairs(n.next);
        n.next = head;
        return n;
    }
}

跟上Reverse Nodes in k-Group.