leetcode

题目：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.

题目要求O(1)的内存消耗和只遍历一遍链表

个人思路：

1、第m-1个结点记为start，第m个结点记为before，第m+1个结点记为current，从current开始向第n个结点前进，每次都把current结点插入到start结点的后面，直到current遍历完成

2、注意一下边界情况即可

代码：

#include <stddef.h>
/*
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)
        {
            return NULL;
        }

        //临时的头结点，仅仅是为了方便接下来的操作
        ListNode *tmpHead = new ListNode(0);
        tmpHead->next = head;

        ListNode *start = tmpHead;
        ListNode *before = NULL;
        ListNode *current = head;
        int pos = 0;
        
        while (++pos != m)
        {
            start = current;
            current = current->next;
        }

        before = current;
        if (++pos <= n)
        {
            current = current->next;
        }

        while (pos <= n)
        {
            before->next = current->next;
            current->next = start->next;
            start->next = current;
            current = before->next;
            ++pos;
        }

        head = tmpHead->next;

        delete tmpHead;

        return head;
    }
};