[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.

这道题在单链表的反转上做了一点点的修改，要求只反转链表的从第m个到第n个结点。分两步走：
1. 定位到第m个结点
2. 进行反转直到第n个结点：没遇到一个结点，就把它插入到第m个结点的前面的位置，然后继续下一个结点。

在操作的过程中需要注意：
1. 需要一个指向第m个结点前驱结点的指针
2. 若m==n，则无需操作。

下面贴上代码：

/** * 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 (m == n)            return head;        ListNode* first = new ListNode(0);        int len = n - m;        first->next = head;        ListNode* p = first;        while (p&&m > 1){            p = p->next;            m--;        }        ListNode* q = p->next;        ListNode* tail = q;        p->next = NULL;        ListNode* r = NULL;        while (q&&len >= 0){            r = q->next;            q->next = p->next;            p->next = q;            q = r;            len--;        }        tail->next = r;        return first->next;    }};