[LeetCode] 25. Reverse Nodes in k-Group 每k个一组翻转链表 Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
Example:
【C++实现LeetCode(25.每k个一组翻转链表)】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
Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

-1->1->2->3->4->5
|||
precur next

-1->3->2->1->4->5
|||
curpre next

class Solution {public:ListNode* reverseKGroup(ListNode* head, int k) {if (!head || k == 1) return head; ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = head; dummy->next = head; for (int i = 1; cur; ++i) {if (i % k == 0) {pre = reverseOneGroup(pre, cur->next); cur = pre->next; } else {cur = cur->next; }}return dummy->next; }ListNode* reverseOneGroup(ListNode* pre, ListNode* next) {ListNode *last = pre->next, *cur = last->next; while(cur != next) {last->next = cur->next; cur->next = pre->next; pre->next = cur; cur = last->next; }return last; }};