[LeetCode]Reverse Linked List II

题目描述：(链接)

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.

解题思路:

头插法

/**
 * 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) {
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *prev = &dummy;
        for (int i = 0; i < m - 1; ++i) {
            prev = prev->next;
        }
        
        ListNode *head2 = prev;
        prev = head2->next;
        ListNode *current = prev->next;
        for (int i = m; i < n; ++i) {
            prev->next = current->next;
            current->next = head2->next;
            head2->next = current;
            current = prev->next;
        }
        
        return dummy.next;
    }
};