题目描述：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
代码如下：
/** * 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(head == nullptr || head->next == nullptr || k < 2) return head; ListNode *next_group = head; for(int i = 0; i < k; i++){ if(next_group) next_group = next_group->next; else return head; } //next_group是未转换的下一组的首地址 //new_next_group是下一组转换之后的首地址 ListNode *new_next_group = reverseKGroup(next_group, k); ListNode *prev = NULL, *cur = head; while(cur != next_group){ ListNode *next = cur->next; cur->next = prev ? prev : new_next_group; prev = cur; cur = next; } return prev; } };
LeetCode 025 Reverse Nodes in k-Group
原文：http://www.cnblogs.com/510602159-Yano/p/4278973.html