25. Reverse Nodes in k-Group

🔴   🔖  递归 链表  🔗 LeetCode

题目

Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2

Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3

Output: [3,2,1,4,5]

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

Follow-up: Can you solve the problem in O(1) extra memory space?

题目大意

按照每 K 个元素翻转的方式翻转链表。如果不满足 K 个元素的就不翻转。

解题思路

这一题是 第 24 题 的加强版，第 24 题是两两相邻的元素，翻转链表。而本题要求的是 k 个相邻的元素，翻转链表，第 24 题相当于是 k = 2 的特殊情况。

代码

/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var reverseKGroup = function (head, k) {
	let node = head;
	// 将 node 指向第 k + 1 个节点
	for (let i = 0; i < k; i++) {
		// 若不足 k 个节点，直接返回 head
		if (!node) {
			return head;
		}
		node = node.next;
	}
	// 翻转 1 ~ k 之间的节点
	let newHead = reverse(head, node);
	// 递归，继续反转后面的节点
	head.next = reverseKGroup(node, k);
	return newHead;
};

// 反转的 first ~ last - 1 之间的节点
/**
 * @param {ListNode} first
 * @param {ListNode} last
 * @return {ListNode}
 */
var reverse = function (first, last) {
	let res = new ListNode(0, first);
	while (first.next != last) {
		let temp = first.next;
		first.next = temp.next;
		temp.next = res.next;
		res.next = temp;
	}
	return res.next;
};

相关题目

相关题目

• 24. 两两交换链表中的节点
• 1721. 交换链表中的节点
• 2074. 反转偶数长度组的节点

- [24. 两两交换链表中的节点](./0024.md)
- [1721. 交换链表中的节点](https://leetcode.com/problems/swapping-nodes-in-a-linked-list)
- [2074. 反转偶数长度组的节点](https://leetcode.com/problems/reverse-nodes-in-even-length-groups)