题目链接：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.

这道题的要求是将链表中m到n位置反转。

链表处理，先加一个空表头，便于处理。然后令l指针移动到m位置前面，令r移动到m位置，这样不断把r后面的节点插入到l后面，直到n位置。

假设m=2，n=4，链表为：
head -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL

加头之后为：
h -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL

令l指针移动到m位置前面，令r移动到m位置，即：
h -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL
          ^    ^
          |    |
          l    r

r后面的节点插入到l后面：
ListNode *temp = l -> next:
h -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL
          ^    ^
          |    |
          l  temp、r 

l -> next = r -> next:
          |-------->|
h -> 0 -> 1    2 -> 3 -> 4 -> 5 -> NULL
          ^    ^
          |    |
          l  temp、r

r -> next = r -> next -> next:
          |-------->|
h -> 0 -> 1    2    3 -> 4 -> 5 -> NULL
          ^    |-------->|
          |    ^
          l    |
             temp、r

l -> next -> next = temp:
          |-------->|
h -> 0 -> 1    2 <- 3    4 -> 5 -> NULL
          ^    |-------->|
          |    ^
          l    |
             temp、r

即：
h -> 0 -> 1 -> 3 -> 2 -> 4 -> 5 -> NULL
          ^         ^
          |         |
          l         r

进而继续把r后面节点插入到l后面，以此类推，进行循环。。。

时间复杂度：O(n)

空间复杂度：O(1)

 1 class Solution
 2 {
 3 public:
 4     ListNode *reverseBetween(ListNode *head, int m, int n)
 5     {
 6         ListNode *h = new ListNode(0);
 7         h -> next = head;
 8         
 9         ListNode *l = h, *r;
10         for(int i = 1; i < m; ++ i)
11             l = l -> next;
12         
13         r = l -> next;
14         for(int i = 1; i < n - m + 1; ++ i)
15         {
16             ListNode *temp = l -> next;
17             l -> next = r -> next;
18             r -> next = r -> next -> next;
19             l -> next -> next = temp;
20         }
21         
22         return h -> next;
23     }
24 };