leetcode92 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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {//边遍历边旋转，目标数目节点旋转完成后，与前后未变部分进行对接
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode *ans=head;
        ListNode *pri=head;
        int count=n-m;
        if(m==1)
            ans=NULL;
        else
            ans=head;
        if(m==1)
        {
            ListNode *rev=pri;
            ListNode *acc=rev->next;
            ListNode *revlast=rev;
            while(count--)
            {
                ListNode *temp=acc;
                acc=acc->next;
                temp->next=rev;
                rev=temp;
            }
            revlast->next=acc;
            ans=rev;
        }
        else
        {
            m--;
            while(--m)
                pri=pri->next;
            
            ListNode *rev=pri->next;
            ListNode *acc=rev->next;
            ListNode *revlast=rev;
            while(count--)
            {
                ListNode *temp=acc;
                acc=acc->next;
                temp->next=rev;
                rev=temp;
            }
            revlast->next=acc;
            pri->next=rev;
        }
        return ans;
    }
};