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:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode swapPairs(ListNode head) {
11         if(head == null || head.next == null){
12             return head;
13         }
14         
15         ListNode dummy = new ListNode(0);
16         dummy.next = head;
17         
18         ListNode mark = dummy;
19         ListNode A;
20         ListNode B;
21         while(mark.next != null && mark.next.next != null){
22             A = mark.next;
23             B = mark.next.next;
24             A.next = B.next;
25             B.next = A;
26             mark.next = B;
27             mark = mark.next.next;
28         }
29         return dummy.next;
30     }
31 }

Recursion解法.

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

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode swapPairs(ListNode head) {
11         if(head == null || head.next == null){
12             return head;
13         }
14         ListNode n = head.next;
15         head.next = swapPairs(n.next);
16         n.next = head;
17         return n;
18     }
19 }

跟上Reverse Nodes in k-Group.