LeetCode::Reverse Linked List II

https://oj.leetcode.com/problems/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 *pCurr = head;
        ListNode *pFirstEnd = NULL;
        while (m > 1 && pCurr != NULL) {
            --m;
            --n;
            pFirstEnd = pCurr;
            pCurr = pCurr->next;
        }
        ListNode *pReverseEnd = pCurr;
        ListNode *pReverseCurr = pCurr;
        ListNode *pNext = NULL;
        pCurr = pCurr->next;

        while (n > 1 && pCurr != NULL) {
            --n;
            pNext = pCurr->next;

            pCurr->next = pReverseCurr;
            pReverseCurr = pCurr;

            pCurr = pNext;
        }

        if (pReverseEnd != NULL) {
            pReverseEnd->next = pCurr;
        }

        if (pFirstEnd != NULL) {
            pFirstEnd->next = pReverseCurr;
        } else {
            head = pReverseCurr;
        }

        return head;
    }
};