Leetcode: 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.

思路分析：
第一次将第一个元素插入到第二个元素后面，将第三个元素插入懂啊第四个后面，依次类推。

C++参考代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution
{
public:
    ListNode *swapPairs(ListNode *head)
    {
        if (!head || !head->next) return head;
        ListNode *previous = nullptr;
        ListNode *current = head;
        ListNode *next = head->next;
        head = head->next;//记录调整后的头节点
        while (next)
        {
            //将current节点插入到next节点后面
            current->next = next->next;
            next->next = current;
            //第一次的时候previous是nullptr
            if (previous) previous->next = next;
            //插入完毕后指针移动下一位
            previous = current;
            current = current->next;
            //最后current变成nullptr，这样直接把next也置为nullptr，退出循环
            if (current) next = current->next;
            else next = nullptr;
        }
        return head;
    }
};

C#参考代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution
{
    public ListNode SwapPairs(ListNode head)
    {
        if (head == null || head.next == null) return head;
        ListNode previous = null;
        ListNode current = head;
        ListNode next = head.next;
        head = head.next;
        while(next != null)
        {
            current.next = next.next;
            next.next = current;
            if (previous != null) previous.next = next;
            previous = current;
            current = current.next;
            if (current == null) next = null;
            else next = current.next;
        }
        return head;
    }
}