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;}}}
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