LeetCode OJ：Reverse Linked List II（反转链表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.

大清早起来就被链表虐哭了啊，看了下别人的，额原来可以这么简单，果然脑子还是转不过来的，实际上是很常见的一个题目，代码很简单，完全不用注释也可以的：

 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 == NULL) return NULL;
13         ListNode * p = head;
14         int i, j;
15         for(i = 1; i < m; ++i){
16             p = p->next;
17         }
18         ListNode * q = p;
19         for(i = m; i < n; ++i){
20             for(j = i; j < n; ++j){
21                 q = q->next;
22             }
23             swap(p->val, q->val);
24             n--;
25             p = p->next;
26             q = p;
27         }
28         return head;
29     }
30 };

注意一下那个swap， swap用的很巧妙。

之后有看到一个大神写出来的，也很简单，贴出来学习一个：

 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         // Start typing your C/C++ solution below
13         // DO NOT write int main() function
14         if (head == NULL)
15             return NULL;
16             
17         ListNode *q = NULL;
18         ListNode *p = head;
19         for(int i = 0; i < m - 1; i++)
20         {
21             q = p;
22             p = p->next;
23         }
24         
25         ListNode *end = p;
26         ListNode *pPre = p;
27         p = p->next;
28         for(int i = m + 1; i <= n; i++)
29         {
30             ListNode *pNext = p->next;
31             
32             p->next = pPre;
33             pPre = p;
34             p = pNext;
35         }
36         
37         end->next = p;
38         if (q)
39             q->next = pPre;
40         else
41             head = pPre;
42         
43         return head;
44     }
45 };

唉唉，经常遇到链表脑子就转不过来，这个还是要多练练啊。

下面贴一个java版本的,方法基本上和第一种差不多,熟悉一下用法：

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode reverseBetween(ListNode head, int m, int n) {
11         int count = 1;
12         ListNode helper = new ListNode(-1);
13         helper.next = head;
14         ListNode p = helper.next;
15         ListNode pPre = helper;
16         while(count != m){
17             p = p.next;
18             pPre = pPre.next;
19             count++;
20         }
21         ListNode midPre = pPre;//第一个节点的位置
22         ListNode tmp = null;
23         while(count != n){
24             tmp = p.next;
25             p.next = pPre;
26             pPre = p;
27             p = tmp;
28             count++;
29         }
30 //        ListNode mid2= p; // 指向第二个节点的位置
31         tmp = p.next;
32         p.next = pPre;
33         midPre.next.next = tmp;
34         midPre.next = p;
35         return helper.next;
36     }
37 }