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
链表题，不解释！
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 void reverse(ListNode *&head, ListNode *&tail) { 12 if (tail == NULL) return; 13 ListNode *pos = head->next, *pre = tail->next, *tmp; 14 head->next = tail; 15 head = pos; 16 while (pos != tail) { 17 tmp = pos->next; 18 pos->next = pre; 19 pre = pos; 20 pos = tmp; 21 } 22 pos->next = pre; 23 tail = head; 24 } 25 26 ListNode *reverseKGroup(ListNode *head, int k) { 27 ListNode *res = new ListNode(-1); 28 res->next = head; 29 ListNode *pre = res, *tail = res; 30 while (tail != NULL) { 31 for (int i = 0; i < k; ++i) { 32 if (tail) tail = tail->next; 33 else return res->next; 34 } 35 reverse(pre, tail); 36 } 37 return res->next; 38 } 39 };
[Leetcode] Reverse Nodes in k-Group,布布扣,bubuko.com
[Leetcode] Reverse Nodes in k-Group
原文：http://www.cnblogs.com/easonliu/p/3663083.html