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：

 /**
  * 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 * p = head;
         int i, j;
         for(i = ; i < m; ++i){
             p = p->next;
         }
         ListNode * q = p;
         for(i = m; i < n; ++i){
             for(j = i; j < n; ++j){
                 q = q->next;
             }
             swap(p->val, q->val);
             n--;
             p = p->next;
             q = p;
         }
         return head;
     }
 };

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：

 /**
  * 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) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         if (head == NULL)
             return NULL;

         ListNode *q = NULL;
         ListNode *p = head;
         for(int i = ; i < m - ; i++)
         {
             q = p;
             p = p->next;
         }

         ListNode *end = p;
         ListNode *pPre = p;
         p = p->next;
         for(int i = m + ; 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;
     }
 };

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：

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public ListNode reverseBetween(ListNode head, int m, int n) {
         int count = 1;
         ListNode helper = new ListNode(-1);
         helper.next = head;
         ListNode p = helper.next;
         ListNode pPre = helper;
         while(count != m){
             p = p.next;
             pPre = pPre.next;
             count++;
         }
         ListNode midPre = pPre;//第一个节点的位置
         ListNode tmp = null;
         while(count != n){
             tmp = p.next;
             p.next = pPre;
             pPre = p;
             p = tmp;
             count++;
         }
 //        ListNode mid2= p; // 指向第二个节点的位置
         tmp = p.next;
         p.next = pPre;
         midPre.next.next = tmp;
         midPre.next = p;
         return helper.next;
     }
 }