[leetcode][list] Reverse Linked List II

题目：

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) {if (NULL == head || NULL == head->next || m == n) return head;ListNode *p = head;ListNode *headNew = head;if (m == 1) headNew = reverseList(head, n);else{for (int i = 1; i < m - 1; ++i){p = p->next;}p->next = reverseList(p->next, n - m + 1);}return headNew;}private:ListNode *reverseList(ListNode *head, int len){if (NULL == head || len == 1) return head;ListNode *headNew = head, *p = head->next;--len;while (len > 0){ListNode *q = p->next;p->next = headNew;headNew = p;p = q;--len;}head->next = p;return headNew;}};