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     }

View Code

SOLUTION 2:

迭代解法：

1. 使用Dummy node保存头节点前一个节点

2. 记录翻转区域的Pre（上一个节点），记录翻转区域的next，或是tail。

3. 翻转特定区域，并不断前移。

有2种写法，后面一种写法稍微简单一点，记录的是翻转区域的下一个节点。

 1 // Solution 2: the iteration version.
 2     public ListNode swapPairs(ListNode head) {
 3         // 如果小于2个元素，不需要任何操作
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7         
 8         ListNode dummy = new ListNode(0);
 9         dummy.next = head;
10         
11         // The node before the reverse area;
12         ListNode pre = dummy;
13         
14         while (pre.next != null && pre.next.next != null) {
15             // The last node of the reverse area;
16             ListNode tail = pre.next.next;
17             
18             ListNode tmp = pre.next;
19             pre.next = tail;
20             
21             ListNode next = tail.next;
22             tail.next = tmp;
23             tmp.next = next;
24             
25             // move forward the pre node.
26             pre = tmp;
27         }
28         
29         return dummy.next;
30     }

View Code

 1 // Solution 3: the iteration version.
 2     public ListNode swapPairs3(ListNode head) {
 3         // 如果小于2个元素，不需要任何操作
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7         
 8         ListNode dummy = new ListNode(0);
 9         dummy.next = head;
10         
11         // The node before the reverse area;
12         ListNode pre = dummy;
13         
14         while (pre.next != null && pre.next.next != null) {
15             ListNode next = pre.next.next.next;
16             
17             ListNode tmp = pre.next;
18             pre.next = pre.next.next;
19             pre.next.next = tmp;
20             
21             tmp.next = next;
22                         
23             // move forward the pre node.
24             pre = tmp;
25         }
26         
27         return dummy.next;
28     }

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/list/SwapPairs3.java