LeetCode: Swap Nodes in Pairs 解题报告

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 1:

递归解法：

1. 先翻转后面的链表，得到新的Next.

2. 翻转当前的2个节点。

3. 返回新的头部。

 1 // Solution 1: the recursion version.
 2     public ListNode swapPairs1(ListNode head) {
 3         if (head == null) {
 4             return null;
 5         }
 6         
 7         return rec(head);
 8     }
 9     
10     public ListNode rec(ListNode head) {
11         if (head == null || head.next == null) {
12             return head;
13         }
14         
15         ListNode next = head.next.next;
16         
17         // 翻转后面的链表
18         next = rec(next);
19         
20         // store the new head.
21         ListNode tmp = head.next;
22         
23         // reverse the two nodes.
24         head.next = next;
25         tmp.next = head;
26         
27         return tmp;
28     }

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