[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.

此题在上题的基础上加了位置要求，只翻转指定区域的链表。由于链表头节点不确定，使用dummy节点。。

1. 由于只翻转指定区域，分析受影响的区域为第m-1个和第n+1个节点
2. 找到第m个节点，使用for循环n-m次，使用经典反转算法中的链表翻转方法
3. 处理第m-1个和第n+1个节点
4. 返回dummy->next

/**
 * 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 (head == nullptr || m > n) 
            return head;
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* node = dummy;
        for (int i = 1; i != m; i++) {
            if (node == nullptr)
                return node;
            else
                node = node->next;
        }  
        ListNode* premNode = node;
        ListNode* mNode = node->next;
        ListNode* nNode = mNode;
        ListNode* postnNode = nNode->next;
        for (int i = m; i != n; i++) {
            ListNode* tmp = postnNode->next;
            postnNode->next = nNode;
            nNode = postnNode;
            postnNode = tmp;
        }
        premNode->next = nNode;
        mNode->next = postnNode;
        
        return dummy->next;
    }
};
// 6 ms