leetcode 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.

For this type of question, two things should be paid attention to: 

1. If the first node is the head, take case of the predecessor of head. 
2. If the first node is the head, take case of the returning value. 

/**
 * 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) {
    // Note: The Solution object is instantiated only once and is reused by each test case.
    if (m == n || head == NULL)
      return head;
    
    ListNode *p = head, *last = NULL, *next = NULL, *lastend = NULL;
    int cur = 1;
    while (p != NULL && cur <= n) {
      next = p->next;
      if (cur == m)
        lastend = last;
      if (cur >= m && cur <= n)
        p->next = last;
            
      last = p;
      p = next;
      ++cur;
    }
    if (lastend == NULL) {
      head->next = p;
      return last;
    }
    else {
      lastend->next->next = p;
      lastend->next = last;
      return head;
    }
  }
};