[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:    ListNode *reverseBetween(ListNode *head, int m, int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        assert(m<=n);                int count = 1;        ListNode* vHead = new ListNode(-1);                vHead->next = head;        ListNode* last = vHead;        ListNode* first = head;                for(int i=1;i<m;i++)        {            last = last->next;            first = first->next;        }                        ListNode* pre = last->next;        ListNode* current = last->next->next;        ListNode* temp;                for(int i=m;i<n;i++)        {            temp = current->next;            current->next = pre;            pre = current;            current = temp;        }                last->next = pre;      //Think Through!        first->next = current;                return vHead->next;        }};