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.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    void reverseList(ListNode* &node1, ListNode* &node2) {                ListNode *pPreNode = NULL;        ListNode *pCurNode = node1;        ListNode *pNextNode = node1->next;                while (pCurNode != node2)        {            pNextNode = pCurNode->next;            pCurNode->next = pPreNode;                        pPreNode = pCurNode;            pCurNode = pNextNode;        }                pCurNode->next = pPreNode;    }    ListNode *reverseBetween(ListNode *head, int m, int n) {                ListNode fakeHead(0);        ListNode *pPreNode = &fakeHead;        ListNode *pCurNode = head;        ListNode *pAfterNode;        ListNode *pBeforeNode;                pPreNode->next = pCurNode;                ListNode *reverseN1;        ListNode *reverseN2;                int step = 1;                while (step <= n && pCurNode)        {            if (step == m)            {                reverseN1 = pCurNode;                pBeforeNode = pPreNode;            }                            if (step == n)            {                reverseN2 = pCurNode;                pAfterNode = pCurNode->next;                break;            }                        pPreNode = pPreNode->next;            pCurNode = pCurNode->next;            step++;        }                if (reverseN1 != reverseN2)        {            reverseList(reverseN1, reverseN2);        }                pBeforeNode->next = reverseN2;        reverseN1->next = pAfterNode;                return fakeHead.next;    }};