反转指定区间内链表

题目：

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given1->2->3->4->5->NULL, m = 2 and n = 4,

return1->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 {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if(head==NULL || head->next==NULL)  return head;
        
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *prev, *p, *q, *last;
        //得到m处的结点
        p = head;
        for(int i=1;i<m;i++)
        {p = p->next;}
        //得到n处的结点
        q = p;
        int x = n-m;
        for(int j=0;j<x;j++)
        { q = q->next;}
        //设定last,断开n处结点与后面的结点
        last = q->next;
        q->next = NULL;
        //设定prev,断开m处结点与前面的结点
        prev = dummy;
        int y = m;
        for(int z=1;z<y;z++)
        { prev = prev->next;}
        prev->next = NULL;
        
        //反转m与n区间间的结点
        ListNode *tmp1 = p->next;
        ListNode *tmp2;
        ListNode *last_new = p;   //lat_new存储反转后的尾结点
        p->next = NULL;
        while(tmp1!=NULL)
        {
            tmp2 = tmp1->next;
            tmp1->next = p;
            p = tmp1;
            tmp1 = tmp2;
        }
        ListNode *fir_new = p;
        
        //重新连接
        prev->next = fir_new;
        last_new->next = last;
        
        return dummy->next;
    }
};