mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 11:08:15 +08:00
168 lines
24 KiB
JSON
168 lines
24 KiB
JSON
{
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"questionId": "25",
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"questionFrontendId": "25",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1096,
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"title": "Reverse Nodes in k-Group",
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"titleSlug": "reverse-nodes-in-k-group",
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"content": "<p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>\n\n<p><code>k</code> 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 <code>k</code> then left-out nodes, in the end, should remain as it is.</p>\n\n<p>You may not alter the values in the list's nodes, only nodes themselves may be changed.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg\" style=\"width: 542px; height: 222px;\" />\n<pre>\n<strong>Input:</strong> head = [1,2,3,4,5], k = 2\n<strong>Output:</strong> [2,1,4,3,5]\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg\" style=\"width: 542px; height: 222px;\" />\n<pre>\n<strong>Input:</strong> head = [1,2,3,4,5], k = 3\n<strong>Output:</strong> [3,2,1,4,5]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the list is <code>n</code>.</li>\n\t<li><code>1 <= k <= n <= 5000</code></li>\n\t<li><code>0 <= Node.val <= 1000</code></li>\n</ul>\n\n<p> </p>\n<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>\n",
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"translatedTitle": "K 个一组翻转链表",
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"translatedContent": "<p>给你一个链表，每 <em>k </em>个节点一组进行翻转，请你返回翻转后的链表。</p>\n\n<p><em>k </em>是一个正整数，它的值小于或等于链表的长度。</p>\n\n<p>如果节点总数不是 <em>k </em>的整数倍，那么请将最后剩余的节点保持原有顺序。</p>\n\n<p><strong>进阶：</strong></p>\n\n<ul>\n\t<li>你可以设计一个只使用常数额外空间的算法来解决此问题吗？</li>\n\t<li><strong>你不能只是单纯的改变节点内部的值</strong>，而是需要实际进行节点交换。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg\" style=\"width: 542px; height: 222px;\" />\n<pre>\n<strong>输入：</strong>head = [1,2,3,4,5], k = 2\n<strong>输出：</strong>[2,1,4,3,5]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg\" style=\"width: 542px; height: 222px;\" />\n<pre>\n<strong>输入：</strong>head = [1,2,3,4,5], k = 3\n<strong>输出：</strong>[3,2,1,4,5]\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>head = [1,2,3,4,5], k = 1\n<strong>输出：</strong>[1,2,3,4,5]\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>head = [1], k = 1\n<strong>输出：</strong>[1]\n</pre>\n\n<ul>\n</ul>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>列表中节点的数量在范围 <code>sz</code> 内</li>\n\t<li><code>1 <= sz <= 5000</code></li>\n\t<li><code>0 <= Node.val <= 1000</code></li>\n\t<li><code>1 <= k <= sz</code></li>\n</ul>\n",
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"code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\n * ListNode(int x) : val(x), next(nullptr) {}\n * ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npublic:\n ListNode* reverseKGroup(ListNode* head, int k) {\n\n }\n};",
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"code": "# Definition for singly-linked list.\n# class ListNode(object):\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution(object):\n def reverseKGroup(self, head, k):\n \"\"\"\n :type head: ListNode\n :type k: int\n :rtype: ListNode\n \"\"\"\n ",
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"code": "# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:",
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"code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * struct ListNode *next;\n * };\n */\n\n\nstruct ListNode* reverseKGroup(struct ListNode* head, int k){\n\n}",
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"code": "/**\n * Definition for singly-linked list.\n * public class ListNode {\n * public int val;\n * public ListNode next;\n * public ListNode(int val=0, ListNode next=null) {\n * this.val = val;\n * this.next = next;\n * }\n * }\n */\npublic class Solution {\n public ListNode ReverseKGroup(ListNode head, int k) {\n\n }\n}",
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"code": "/**\n * Definition for singly-linked list.\n * function ListNode(val, next) {\n * this.val = (val===undefined ? 0 : val)\n * this.next = (next===undefined ? null : next)\n * }\n */\n/**\n * @param {ListNode} head\n * @param {number} k\n * @return {ListNode}\n */\nvar reverseKGroup = function(head, k) {\n\n};",
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"code": "# Definition for singly-linked list.\n# class ListNode\n# attr_accessor :val, :next\n# def initialize(val = 0, _next = nil)\n# @val = val\n# @next = _next\n# end\n# end\n# @param {ListNode} head\n# @param {Integer} k\n# @return {ListNode}\ndef reverse_k_group(head, k)\n\nend",
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"code": "/**\n * Definition for singly-linked list.\n * public class ListNode {\n * public var val: Int\n * public var next: ListNode?\n * public init() { self.val = 0; self.next = nil; }\n * public init(_ val: Int) { self.val = val; self.next = nil; }\n * public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }\n * }\n */\nclass Solution {\n func reverseKGroup(_ head: ListNode?, _ k: Int) -> ListNode? {\n\n }\n}",
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"code": "/**\n * Definition for singly-linked list.\n * type ListNode struct {\n * Val int\n * Next *ListNode\n * }\n */\nfunc reverseKGroup(head *ListNode, k int) *ListNode {\n\n}",
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"code": "/**\n * Definition for singly-linked list.\n * class ListNode(_x: Int = 0, _next: ListNode = null) {\n * var next: ListNode = _next\n * var x: Int = _x\n * }\n */\nobject Solution {\n def reverseKGroup(head: ListNode, k: Int): ListNode = {\n\n }\n}",
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"code": "/**\n * Example:\n * var li = ListNode(5)\n * var v = li.`val`\n * Definition for singly-linked list.\n * class ListNode(var `val`: Int) {\n * var next: ListNode? = null\n * }\n */\nclass Solution {\n fun reverseKGroup(head: ListNode?, k: Int): ListNode? {\n\n }\n}",
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"code": "// Definition for singly-linked list.\n// #[derive(PartialEq, Eq, Clone, Debug)]\n// pub struct ListNode {\n// pub val: i32,\n// pub next: Option<Box<ListNode>>\n// }\n//\n// impl ListNode {\n// #[inline]\n// fn new(val: i32) -> Self {\n// ListNode {\n// next: None,\n// val\n// }\n// }\n// }\nimpl Solution {\n pub fn reverse_k_group(head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {\n\n }\n}",
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