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。

 // Solution 1: the recursion version.

     public ListNode swapPairs1(ListNode head) {

         if (head == null) {

             return null;

         }

         return rec(head);

     }

     public ListNode rec(ListNode head) {

         if (head == null || head.next == null) {

             return head;

         }

         ListNode next = head.next.next;

         // 翻转后面的链表

         next = rec(next);

         // store the new head.

         ListNode tmp = head.next;

         // reverse the two nodes.

         head.next = next;

         tmp.next = head;

         return tmp;

     }

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。

 // Solution 2: the iteration version.

     public ListNode swapPairs(ListNode head) {

         // 如果小于2个元素，不需要任何操作

         if (head == null || head.next == null) {

             return head;

         }

         ListNode dummy = new ListNode(0);

         dummy.next = head;

         // The node before the reverse area;

         ListNode pre = dummy;

         while (pre.next != null && pre.next.next != null) {

             // The last node of the reverse area;

             ListNode tail = pre.next.next;

             ListNode tmp = pre.next;

             pre.next = tail;

             ListNode next = tail.next;

             tail.next = tmp;

             tmp.next = next;

             // move forward the pre node.

             pre = tmp;

         }

         return dummy.next;

     }

 // Solution 3: the iteration version.

     public ListNode swapPairs3(ListNode head) {

         // 如果小于2个元素，不需要任何操作

         if (head == null || head.next == null) {

             return head;

         }

         ListNode dummy = new ListNode(0);

         dummy.next = head;

         // The node before the reverse area;

         ListNode pre = dummy;

         while (pre.next != null && pre.next.next != null) {

             ListNode next = pre.next.next.next;

             ListNode tmp = pre.next;

             pre.next = pre.next.next;

             pre.next.next = tmp;

             tmp.next = next;

             // move forward the pre node.

             pre = tmp;

         }

         return dummy.next;

     }

GITHUB:

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

巴特西

LeetCode: Swap Nodes in Pairs 解题报告

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