LeetCode: Reverse Linked List II

没做出来，看网上答案，这题难度在于编程

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!head) return NULL;
        ListNode *q = NULL;
        ListNode *p = head;
        for (int i = 0; i < m-1; i++) {
            q = p;
            p = p->next;
        }
        ListNode *end = p;
        ListNode *pPre = p;
        p = p->next;
        for (int i = m+1; i <= n; i++) {
            ListNode *pNext = p->next;
            p->next = pPre;
            pPre = p;
            p = pNext;
        }
        end->next = p;
        if (q) q->next = pPre;
        else head = pPre;
        return head;
    }
};

 C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode ReverseBetween(ListNode head, int m, int n) {
        if (head == null) return null;
        ListNode q = null, p = head;
        for (int i = 0; i < m-1; i++) {
            q = p;
            p = p.next;
        }
        ListNode end = p, pPre = p;
        p = p.next;
        for (int i = m+1; i <= n; i++) {
            ListNode pNext = p.next;
            p.next = pPre;
            pPre = p;
            p = pNext;
        }
        end.next = p;
        if (q != null) q.next = pPre;
        else head = pPre;
        return head;
    }
}