problem：
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.
thinking:
O（n）算法
code:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
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)
q->next = pPre;
else
head = pPre;
return head;
}
};leetcode || 92、Reverse Linked List II
原文地址：http://blog.csdn.net/hustyangju/article/details/45059613
