leetcode-problemset/leetcode-cn/originData/reverse-nodes-in-k-group.json

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        "question": {
            "questionId": "25",
            "questionFrontendId": "25",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1096,
            "title": "Reverse Nodes in k-Group",
            "titleSlug": "reverse-nodes-in-k-group",
            "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&#39;s nodes, only nodes themselves may be changed.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">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 class=\"example\">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>&nbsp;</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 &lt;= k &lt;= n &lt;= 5000</code></li>\n\t<li><code>0 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>\n",
            "translatedTitle": "K 个一组翻转链表",
            "translatedContent": "<p>给你链表的头节点 <code>head</code> ，每&nbsp;<code>k</code><em>&nbsp;</em>个节点一组进行翻转，请你返回修改后的链表。</p>\n\n<p><code>k</code> 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是&nbsp;<code>k</code><em>&nbsp;</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>你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。</p>\n\n<p>&nbsp;</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\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg\" style=\"width: 542px; height: 222px;\" /></p>\n\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>&nbsp;</p>\n<strong>提示：</strong>\n\n<ul>\n\t<li>链表中的节点数目为 <code>n</code></li>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>\n\t<li><code>0 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以设计一个只用 <code>O(1)</code> 额外内存空间的算法解决此问题吗？</p>\n\n<ul>\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": "/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     int val;\n *     ListNode next;\n *     ListNode() {}\n *     ListNode(int val) { this.val = val; }\n *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }\n * }\n */\nclass Solution {\n    public 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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                    "lang": "C",
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                    "code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n *     int val;\n *     struct ListNode *next;\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": "/**\n * Definition for singly-linked list.\n * class ListNode {\n *     val: number\n *     next: ListNode | null\n *     constructor(val?: number, next?: ListNode | null) {\n *         this.val = (val===undefined ? 0 : val)\n *         this.next = (next===undefined ? null : next)\n *     }\n * }\n */\n\nfunction reverseKGroup(head: ListNode | null, k: number): ListNode | null {\n    \n};",
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                    "code": "/**\n * Definition for a singly-linked list.\n * class ListNode {\n *     public $val = 0;\n *     public $next = null;\n *     function __construct($val = 0, $next = null) {\n *         $this->val = $val;\n *         $this->next = $next;\n *     }\n * }\n */\nclass Solution {\n\n    /**\n     * @param ListNode $head\n     * @param Integer $k\n     * @return ListNode\n     */\n    function reverseKGroup($head, $k) {\n\n    }\n}",
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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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                    "lang": "Kotlin",
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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": "/**\n * Definition for singly-linked list.\n * class ListNode {\n *   int val;\n *   ListNode? next;\n *   ListNode([this.val = 0, this.next]);\n * }\n */\nclass Solution {\n  ListNode? reverseKGroup(ListNode? head, int k) {\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": "# 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": "// 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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                    "code": "; Definition for singly-linked list:\n#|\n\n; val : integer?\n; next : (or/c list-node? #f)\n(struct list-node\n  (val next) #:mutable #:transparent)\n\n; constructor\n(define (make-list-node [val 0])\n  (list-node val #f))\n\n|#\n\n(define/contract (reverse-k-group head k)\n  (-> (or/c list-node? #f) exact-integer? (or/c list-node? #f))\n  )",
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                    "lang": "Erlang",
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                    "code": "%% Definition for singly-linked list.\n%%\n%% -record(list_node, {val = 0 :: integer(),\n%%                     next = null :: 'null' | #list_node{}}).\n\n-spec reverse_k_group(Head :: #list_node{} | null, K :: integer()) -> #list_node{} | null.\nreverse_k_group(Head, K) ->\n  .",
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap 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