Reverse Nodes in k-Group

Reverse Nodes in k-Group 

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

 

解题方法：

采用分段的方法，K个结点为一段，进行Reverse操作，直到链表的末尾，或者剩下的结点数不足K个。

 

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *reverseKGroup(ListNode *head, int k){ListNode *p=head;if(head==NULL ||head->next==NULL ||k<=1)return head;int len=0;while(p){len++;p=p->next;}if(k>len)return head;int cnt=k;ListNode *q=head;p=NULL;while(q && cnt>0){ListNode* pNext=q->next;q->next=p;p=q;q=pNext;cnt--;}    if(len-k>=k)head->next=reverseKGroup(q,k);elsehead->next=q;return p;}};