题目描述：
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
思路：
1. 使用栈来保存m到n之间的数字，其余元素使用队列保存
2. 在[m,n]区间外时，循环弹出栈内元素到链表
3. 在[m,n]区间内，先循环弹出队列元素到链表，再创建栈，最后需要判断当前head是否为空
实现代码：
/**
* 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 ReverseBetween(ListNode head, int m, int n) {
var stack = new Stack<int>();
var q = new Queue<int>();
ListNode node = null;
var c = 1;
ListNode newNode = null;
while(head != null)
{
if(c >= m && c <=n){
while(q.Count > 0){
var first = MoveNext(ref node , q.Dequeue());
if(first){
newNode = node;
}
}
while(c >= m && c<= n){
stack.Push(head.val);
head = head.next;
c++;
}
if(head == null){
while(stack.Count > 0){
var first = MoveNext(ref node , stack.Pop());
if(first){
newNode = node;
}
}
}
}
else{
while(stack.Count > 0){
var first = MoveNext(ref node , stack.Pop());
if(first){
newNode = node;
}
}
var f = MoveNext(ref node , head.val);
if(f)
{
newNode = node;
}
head = head.next;
c++;
}
}
return newNode;
}
private bool MoveNext(ref ListNode n , int val){
if(n == null){
n = new ListNode(val);
return true;
}
else{
n.next = new ListNode(val);
n = n.next;
return false;
}
}
}