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;
}
*/
};