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.

分析：在纸上画图，弄清楚插入node的过程。

运行时间：4ms

ATTENTION:

为什么return head会报错？

为什么要构建ListNode型的dummy而不是ListNode*型的dummy？（指针的深复制与浅复制）

/**
 * 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) {
        if(!head) return head;
        if(!head->next || m == n) return head;
        
        ListNode dummy(-1);
        dummy.next = head;
        ListNode* pre_m, *beforeCur = &dummy;
        ListNode* current = head;
        
        for(int i = 1; i <= n; i++){
            if(i == m) pre_m = beforeCur;
            
            if(i > m && i <= n){
                beforeCur->next = current->next;
                current->next = pre_m->next;
                pre_m->next = current;
                current = beforeCur;
            }
            beforeCur = current;
            current = current->next;
        }
        return dummy.next;
    }
};