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
思路：
链表反向的变形，反向依然是从前到后逐一进行。
题解：
/** * 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) { if(k==1) return head; ListNode *p, *q, *pre; bool flag = true; int count = 1; p=head; q=head; pre=NULL; while(q!=NULL) { count = 1; for( ;q!=NULL && count<k;count++) q = q->next; if(q==NULL) break; if(flag) { flag = false; head = q; } q = q->next; ListNode *p1 = p; ListNode *tmp = NULL; ListNode *pre1 = NULL; while(p!=q) { tmp = p->next; p->next = pre1; pre1 = p; p = tmp; } p1->next = p; if(pre) pre->next = pre1; pre = p1; } return head; } };
[leetcode] Reverse Nodes in k-Group
原文：http://www.cnblogs.com/jiasaidongqi/p/4242930.html
