【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.

思路：局部反转链表，然后在合并。

/**
 * 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 == NULL || head->next == NULL)
        {
            return head;
        }
        ListNode *sg = new ListNode(-1);
        sg->next = head;
        head = sg;
        ListNode *pre = head;
        for(int i = 1; i < m; i++)
        {
            pre = pre->next;
        }
        ListNode *cur = pre->next;
        ListNode *cur_next = cur->next;
        if(n - m > 0)
        {
            int step = n - m;
            while(step > 0 && cur_next != NULL)
            {
                ListNode *temp = cur_next->next;
                cur_next->next = cur;
                cur = cur_next;
                cur_next = temp;
                step--;
            }
            ListNode *temp = pre->next;
            pre->next = cur;
            temp->next = cur_next;
        }
        head = head->next;
        delete sg;
        return head;
    }
};