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.

思路：只要会了reverse k-group，反转链表再也不用愁了呢！

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || m == n) return head;
        ListNode prenode(-1);
        prenode.next = head;
        ListNode *pre = &prenode, *cur, *nex;
        int count = m - 1;
        while (count && (pre = pre->next))
            count--;
        cur = pre->next;
        nex = cur->next;
        for (int i = 1; i < n - m + 1; i++)
        {
            cur->next = nex->next;
            nex->next = pre->next;
            pre->next = nex;
            nex = cur->next;
        }
        return prenode.next;
    }
};