LeetCode: Reverse Linked List
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.
地址：https://oj.leetcode.com/problems/reverse-linked-list-ii/
算法：先找到第m个节点p以及其前趋pre，然后把p记下，因为p是逆置链表部分的最后一个节点。然后从p开始遍历到第n个节点，并把节点依次插入pre后面，注意处理pre为空，也就是m=1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 if(!head) return NULL;
13 ListNode *pre = NULL;
14 ListNode *p = head;
15 int i = 1;
16 while(p && i != m){
17 pre = p;
18 p = p->next;
19 ++i;
20 }
21 if(!p) return head;
22 ListNode *first_reverse_node = p;
23 if(pre){
24 pre->next = NULL;
25 }
26 ListNode *q = p;
27 while(q && i <= n){
28 q = p->next;
29 if(pre){
30 p->next = pre->next;
31 pre->next = p;
32 }else{
33 if(p == head)
34 head->next = NULL;
35 else{
36 p->next = head;
37 head = p;
38 }
39 }
40 p = q;
41 ++i;
42 }
43 first_reverse_node->next = q;
44 return head;
45 }
46 };