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.

 

C++实现代码：

#include
     
      #include
      
       using namespace std;//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)            return NULL;        ListNode *p=head;        ListNode *pre=head;        ListNode *qpre=NULL;        ListNode *q=NULL;        int count=0;        while(p)        {            count++;            if(count==m)            {                qpre=pre;                q=p;            }            else if(count==n)                break;            pre=p;            p=p->next;        }        cout<
       
        val<
        
         val<
         
          next;        ListNode *qq=NULL;        p->next=NULL;        p=q->next;        cout<
          
           val<
           
            next=pp; while(p) { qq=p->next; cout<
            
             val<
             
              next=q; q=p; cout<
              
               val<
               
                val<
                
                 next=q; else head=q; return head; } void createList(ListNode *&head) { ListNode *p=NULL; int i=0; int arr[10]= { 9,8,5,4,4,3,3,3,2,1}; for(i=0; i<3; i++) { if(head==NULL) { head=new ListNode(arr[i]); if(head==NULL) return; } else { p=new ListNode(arr[i]); p->next=head; head=p; } } }};int main(){ Solution s; ListNode *L=NULL; s.createList(L); ListNode *head=L; while(head) { cout<
                 
                  val<<" "; head=head->next; } cout<
                  
                   val<<" "; L=L->next; }}