25 Reverse Nodes in k-Group

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分类专栏： leetcode 文章标签： leetcode

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25 Reverse Nodes in k-Group

链接：https://leetcode.com/problems/reverse-nodes-in-k-group/
问题描述：
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Hide Tags Linked List

这个提就是让链表中相连的k个元素顺序倒过来。那么可以用累加器，当收集了k个元素之后就求List的逆序。

class Solution {
public:
    ListNode* reverseList(ListNode* head)
    {

        ListNode *t=new ListNode(0),*cur=NULL;
        t->next=head;
        cur=head;
        head=head->next;
        cur->next=NULL;
        while(head)
        {
            t->next=head;
            head=head->next;
            t->next->next=cur;
            cur=t->next;

        }
        cur=t->next;
        delete t;
        return cur;
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
         if(head==NULL||head->next==NULL)
                return head;
          ListNode*t=head,*pre,*result,*last;
          int length=0;
          while (t)
          {
            length++;
            t=t->next;
          }
          t=head;
          last=head;
          if(k>length)
          return head;

          for(int i=0;i<length;i++)
          {
              pre=t;
              t=t->next;
              if((i+1)%k==0)
              {
                  pre->next=NULL;
                  if(i+1==k)
                    result=reverseList(head);
                  else
                  {
                      last->next=reverseList(head); 
                      last=head;
                  }
                  head=t;
              }

          }
          last->next=head;
          return result;
    }
};