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:
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.
这道题让我们以每k个为一组来翻转链表，实际上是把原链表分成若干小段，然后分别对其进行翻转，那么肯定总共需要两个函数，一个是用来分段的，一个是用来翻转的，我们就以题目中给的例子来看，对于给定链表1->2->3->4->5，一般在处理链表问题时，我们大多时候都会在开头再加一个dummy node，因为翻转链表时头结点可能会变化，为了记录当前最新的头结点的位置而引入的dummy node，那么我们加入dummy node后的链表变为-1->1->2->3->4->5，如果k为3的话，我们的目标是将1,2,3翻转一下，那么我们需要一些指针，pre和next分别指向要翻转的链表的前后的位置，然后翻转后pre的位置更新到如下新的位置：
-1->1->2->3->4->5 | | pre next -1->3->2->1->4->5 | | pre next
以此类推，只要next走过k个节点，就可以调用翻转函数来进行局部翻转了，代码如下所示：
解法一：
class Solution { public: ListNode *reverseKGroup(ListNode *head, int k) { if (!head || k == 1) return head; ListNode *dummy = new ListNode(-1); ListNode *pre = dummy, *cur = head; dummy->next = head; int i = 0; while (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; ListNode *cur = last->next; while(cur != next) { last->next = cur->next; cur->next = pre->next; pre->next = cur; cur = last->next; } return last; } };
我们也可以在一个函数中完成，我们首先遍历整个链表，统计出链表的长度，然后如果长度大于等于k，我们开始交换节点，当k=2时，每段我们只需要交换一次，当k=3时，每段需要交换2此，所以i从1开始循环，注意交换一段后更新pre指针，然后num自减k，直到num<k时循环结束，参见代码如下：
解法二：
class Solution { public: ListNode* reverseKGroup(ListNode* head, int k) { ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = pre; dummy->next = head; int num = 0; while (cur = cur->next) ++num; while (num >= k) { cur = pre->next; for (int i = 1; i < k; ++i) { ListNode *t = cur->next; cur->next = t->next; t->next = pre->next; pre->next = t; } pre = cur; num -= k; } return dummy->next; } };
我们也可以使用递归来做，我们用head记录每段的开始位置，cur记录结束位置的下一个节点，然后我们调用reverse函数来将这段翻转，然后得到一个new_head，原来的head就变成了末尾，这时候后面接上递归调用下一段得到的新节点，返回new_head即可，参见代码如下：
解法三：
class Solution { public: ListNode* reverseKGroup(ListNode* head, int k) { ListNode *cur = head; for (int i = 0; i < k; ++i) { if (!cur) return head; cur = cur->next; } ListNode *new_head = reverse(head, cur); head->next = reverseKGroup(cur, k); return new_head; } ListNode* reverse(ListNode* head, ListNode* tail) { ListNode *pre = tail; while (head != tail) { ListNode *t = head->next; head->next = pre; pre = head; head = t; } return pre; } };
类似题目：
参考资料：
https://leetcode.com/problems/reverse-nodes-in-k-group/
https://leetcode.com/problems/reverse-nodes-in-k-group/discuss/11435/C%2B%2B-Elegant-and-Small
