LeetCode-- Reverse Linked List II

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;        }    }    }