題目：
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.
原題：https://leetcode.com/problems/swap-nodes-in-pairs/
分析：題意要求將給定的鏈表，相鄰的結點交換位置，並返回變化後的鏈表。
基本思路利用指針移位法，注意考慮特殊情況，如鏈表為空，鏈表長度為0，1，2等。
如下圖是思路的示意：
![這裡寫圖片描述](https://www.aspphp.online/bianchen/UploadFiles_4619/201701/2017012112361107.png "\")
不同顏色表示不同次的循環過程。<喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">vcD4NCjxwPmphdmG0+sLrIEFjY2VwdGVkOjwvcD4NCjxwcmUgY2xhc3M9"brush:java;">
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode p;
ListNode q;
ListNode k;
p = head;
q = p.next;
k = q.next;
head = head.next;
while (q != null) {
q.next = p;
if(k != null && k.next != null){
p.next = k.next;
p = k;
q = k.next;
k = k.next.next;
}else{
p.next = k;
q = null;
k = null;
}
/*p = k;
if(k != null){
q = k.next;
}else{
q = null;
}
if(k != null && k.next != null){
k = k.next.next;
}else{
k = null;
}*/
}
return head;
}
}
