leetcode[92]Reverse Linked List 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.

/**
* 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)
{
ListNode *newh=new ListNode(0);
newh->next=head;
ListNode *p=newh;
ListNode *pre=newh, *left, *right, *after;
int i=0;
while (p&&i<m)
{
pre=p;
p=p->next;
left=p;
right=left;
i++;
}
p=p->next;
after=p;
for (i=m;i<n;i++)
{
p=p->next;
right->next=after->next;
pre->next=after;
after->next=left;
left=after;
after=p;
}
return newh->next;
}
/*
ListNode *reverseBetween(ListNode *head, int m, int n)
{
ListNode *newh=new ListNode(0);
newh->next=head;
ListNode *p=newh->next;
int i=1;
vector<ListNode *> vec;
while(p&&i<=n)
{
if(i>=m&&i<=n)
{
vec.push_back(p);
}
i++;
p=p->next;
}
for (int j=0;j<(n-m+1)/2;j++)
{
int tmp=vec[j]->val;
vec[j]->val=vec[(n-m)-j]->val;
vec[(n-m)-j]->val=tmp;
}
return newh->next;
}
*/
};