[Leetcode] Reverse Linked List II

Problem Description:

Reverse a linked list from position m toN. 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.

Basic ideas:

The idea is simple, flip the list of points to note the pointer can be.

Code:

Public ListNode Reverse (listnode head,int N)  //java  {listnode   tmp = head;int Step = 1;tmp = Tmp.next; ListNode tmphead = head; ListNode p = null; ListNode pre = Head;while (Step < n) {p = Tmp.next;tmp.next = Head;head = Tmp;pre.next = P;//head.next = Null;tmp = P;st ep++;} Tmphead.next = P;return head;  }  Public ListNode Reversebetween (listnode head, int m, int n) {        if (head = = NULL | | head.next = = NULL | | m==n)        return Head;        ListNode result = new ListNode (0);        Result.next = head;        ListNode p = head;        ListNode pre =result;        int pos = 1;        while (pos! = m) {        pre = P;        p = p.next;        pos++;                }        ListNode tmphead = reverse (P, n-m+1);        Pre.next = Tmphead;        return result.next;    }

[Leetcode] Reverse Linked List II