LeetCode92 Reverse Linked List II

题目：

Reverse a linked list from position m to n. Do it in-place and in one-pass. （Medium）

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.

分析：

链表部分翻转。还是画一个图就比较清晰，注意保存翻转部分的前一个的位置（示例中的1）和翻转的第一个位置（翻转后最翻转部分的最后一个，示例中的2）用来链接链表。

使用dummy node，统一处理当链表头也被翻转的情况。

代码：

1 /**
2  * Definition for singly-linked list.
3  * struct ListNode {
4  *     int val;
5  *     ListNode *next;
6  *     ListNode(int x) : val(x), next(NULL) {}
7  * };
8  */
9 class Solution {
10 public:
11     ListNode* reverseBetween(ListNode* head, int m, int n) {
12         ListNode dummy(0);
13         dummy.next = head;
14         head = &dummy;
15         ListNode* temp = head;
16         for (int i = 1; i < m; ++i) {
17             temp = temp -> next;
18         }
19         ListNode* reverseEnd = nullptr;
20         ListNode* reverseBegin = temp -> next;
21         ListNode* p = reverseBegin;
22         for (int i = 0; i <= n - m; ++i) {
23             ListNode* next = p -> next;
24             p -> next = reverseEnd;
25             reverseEnd = p;
26             p = next;
27         }
28         temp -> next = reverseEnd;
29         reverseBegin -> next = p;
30         return dummy.next;
31     }
32 };