35：Reverse Nodes in k-Group

题目：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

解题代码如下：

class Solution {public:        ListNode* reverseKGroup(ListNode* head, int k) {                ListNode dummy{-1};                dummy.next = head;                ListNode *prev = &dummy, *cur = &dummy;                // cur 指向第 k 个节点，prev 指向第一个节点之前的节点                for (int i = 0; cur && i != k; ++i)                        cur = cur -> next;                if (!cur) return head;                for (int i = 0; ; ++i) {                        // 每隔 k 个节点就颠倒这 k 个节点的顺序                        if (i % k == 0) {                                ListNode* tmp = prev -> next;                                reverseList(prev -> next, cur -> next);                                prev -> next = cur;                                cur = tmp;                        }                        if (!cur -> next) break;                        prev = prev -> next;                        cur = cur -> next;                }                return dummy.next;        }private:        //颠倒[first, last)内的节点的顺序        ListNode* reverseList(ListNode* first, ListNode* last) {                ListNode dummy{-1};                dummy.next = first;                while (first -> next != last) {                        ListNode* tmp = first -> next;                        first -> next = tmp -> next;                        tmp -> next = dummy.next;                        dummy.next = tmp;                }                return dummy.next;        }};