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 = 1; 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 = 0; i < m - 1; i++)
{
q = p;
p = p->next;
}

ListNode *end = p;
ListNode *pPre = p;
p = p->next;
for(int i = m + 1; 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;
}
}