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.
题意：反转链表，反转指定位置的链表。
思路：参考这里，主要是先找到要反转链表的开始反转的结点，然后一个个的反转，最后和其两边的部分进行连接。
代码：
public ListNode reverseBetween(ListNode head, int m, int n) { if(head==null) return null; ListNode q = null; ListNode p = head; for(int i=0;i<m-1;i++){ q = p; p = p.next; } ListNode end = p; ListNode pPre = p; p = p.next; for(int i=m+1;i<=n;i++){ ListNode pNext = p.next; p.next = pPre; pPre = p; p = pNext; } end.next = p; if(q!=null) q.next = pPre; else head = pPre; return head; }
LeetCode(92):Reverse Linked List II
原文：http://www.cnblogs.com/Lewisr/p/5140214.html