[LeetCode] Reverse Linked List II

[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.

This problem is slightly modified on the reverse of a single-chain table, requiring that only the chain table be reversed from the m to the n node. Take two steps:
1. Locate the m Node
2. Reverse until the nth node: If a node is not met, insert it to the front of the m node and continue to the next node.

Note the following during the operation:
1. A pointer pointing to the first m node is required.
2. If m = n, no operation is required.

Paste the code below:

/** * 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;    }};