LeetCode-92.Reverse Linked List II

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */

使用栈：（低效）

public ListNode ReverseBetween(ListNode head, int m, int n) 
    {
        if(head.next==null||m==n)
            return head;
         
          ListNode pre = new ListNode(0);
            pre.next = head;
            ListNode curNode = pre;
            Stack<ListNode> stack = new Stack<ListNode>();
            for (int i = 0; i < m; i++)
            {
                curNode = curNode.next;
            }
            for (int i = m; i <= n; i++)
            {
                stack.Push(curNode);
                curNode = curNode.next;
            }
            ListNode leftNode = curNode;
            curNode = pre;
            for (int i = 0; i < m - 1; i++)
            {
                curNode = curNode.next;
            }

            while (stack.Count!=0)
            {
                curNode.next = stack.Pop();
                curNode = curNode.next;
            }
            curNode.next = leftNode;
            return pre.next;
    }

不用栈

public ListNode ReverseBetween(ListNode head, int m, int n) 
    {
       if (head.next == null || m == n)
                return head;

            ListNode pre = new ListNode(0);
            pre.next = head;
            ListNode curNode = pre;
            for (int i = 0; i < m - 1; i++)
            {
                curNode = curNode.next;
            }
            ListNode first = curNode.next;
            ListNode next = first.next;
            for (int i = m; i < n; i++)
            {
                first.next = next.next;
                next.next = curNode.next;
                curNode.next = next;
                next = first.next;
            }
            return pre.next;
    }