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.

這道題自己想的方法沒有寫出來，然后看了別人的解法。發現意思跟我的思路差不多，但細節很多地方我確實想不出來。linkedlist的題目很多細節問題，哪個該連哪個，對我來說現在很容易錯或者陷入混亂。
該解法實際上仍然是維護三個pointers：prev, curt, temp.

以節點為偶數個時舉例：

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當節點數為偶數的時候，是curt == null退出循環；當節點數為奇數的時候，是curt.next == null退出循環。

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null){
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode curt = head;
        while (curt != null && curt.next != null){
            ListNode temp = curt.next.next;
            curt.next.next = curt;
            prev.next = curt.next;
            curt.next = temp;
            prev = curt;
            curt = curt.next;
        }
        return dummy.next;
    }
}