题目连接：

/*题意：将链表中第m到n个结点翻转 *//** *思路：为更好处理表头和第m个结点，引入root结点，同时记录 *      第m-1个结点。从第m个结点开始遍历至第n个结点，将已经 *      遍历过的结点插入在第m-1个结点后，并保证第m个结点的next *      指向遍历结点的next，以避免链表断裂 */class Solution {public:    ListNode *reverseBetween(ListNode *head, int m, int n) {        ListNode *root = new ListNode(0);        root->next = head;        ListNode *mNode = new ListNode(0);        ListNode *pre = root , *curr = head;        for(int i = 1; i <= n; i ++) {            if(i == m) mNode = curr;  //记录第m个结点            if(i < m) pre = curr;     //记录第m-1个结点            ListNode *next = curr->next;            if(i > m && i <= n) {                mNode->next = next;   //避免链表断裂                curr->next = pre->next;                pre->next = curr;            }            curr = next;        }        return root->next;    }};