题目描述：
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.
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就是成对的交换链表的相邻元素。
比较直接的思路是两次遍历：
第一次遍历使用数据结构存链表节点对（本例使用的是sector类）
第二次遍历节点生成新链表并返回其头指针即可。
/**
* 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;
}
var queue = new Queue<Sector>();
var h = head;
while(h != null){
if(h.next == null){
queue.Enqueue(new Sector(h.val, null));
h = h.next;
}
else{
// there is a lot ,continue make queue
queue.Enqueue(new Sector(h.val, h.next.val));
h = h.next;
h = h.next;
}
}
ListNode result = null;
ListNode r = null;
while(queue.Count > 0){
var s = queue.Dequeue();
if(result == null){
result = s.node2;
r = result;
result.next = s.node1;
result = result.next;
}else{
if(s.node2 != null){
result.next = s.node2;
result = result.next;
result.next = s.node1;
result = result.next;
}
else{
result.next = s.node1;
result = result.next;
}
}
}
return r;
}
public class Sector{
public ListNode node1;
public ListNode node2;
public Sector(int? v1, int? v2)
{
if(v1.HasValue){
node1 = new ListNode(v1.Value);
}
if(v2.HasValue){
node2 = new ListNode(v2.Value);
}
}
}
}