leetcode-problemset/leetcode-cn/originData/middle-of-the-linked-list.json

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        "question": {
            "questionId": "908",
            "questionFrontendId": "876",
            "categoryTitle": "Algorithms",
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            "title": "Middle of the Linked List",
            "titleSlug": "middle-of-the-linked-list",
            "content": "<p>Given the <code>head</code> of a singly linked list, return <em>the middle node of the linked list</em>.</p>\n\n<p>If there are two middle nodes, return <strong>the second middle</strong> node.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/07/23/lc-midlist1.jpg\" style=\"width: 544px; height: 65px;\" />\n<pre>\n<strong>Input:</strong> head = [1,2,3,4,5]\n<strong>Output:</strong> [3,4,5]\n<strong>Explanation:</strong> The middle node of the list is node 3.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/07/23/lc-midlist2.jpg\" style=\"width: 664px; height: 65px;\" />\n<pre>\n<strong>Input:</strong> head = [1,2,3,4,5,6]\n<strong>Output:</strong> [4,5,6]\n<strong>Explanation:</strong> Since the list has two middle nodes with values 3 and 4, we return the second one.\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 in the range <code>[1, 100]</code>.</li>\n\t<li><code>1 &lt;= Node.val &lt;= 100</code></li>\n</ul>\n",
            "translatedTitle": "链表的中间结点",
            "translatedContent": "<p>给定一个头结点为 <code>head</code> 的非空单链表，返回链表的中间结点。</p>\n\n<p>如果有两个中间结点，则返回第二个中间结点。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>[1,2,3,4,5]\n<strong>输出：</strong>此列表中的结点 3 (序列化形式：[3,4,5])\n返回的结点值为 3 。 (测评系统对该结点序列化表述是 [3,4,5])。\n注意，我们返回了一个 ListNode 类型的对象 ans，这样：\nans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next = NULL.\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>[1,2,3,4,5,6]\n<strong>输出：</strong>此列表中的结点 4 (序列化形式：[4,5,6])\n由于该列表有两个中间结点，值分别为 3 和 4，我们返回第二个结点。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>给定链表的结点数介于 <code>1</code> 和 <code>100</code> 之间。</li>\n</ul>\n",
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                    "name": "Linked List",
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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* middleNode(ListNode* head) {\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 middleNode(ListNode head) {\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 middleNode(self, head):\n        \"\"\"\n        :type head: ListNode\n        :rtype: ListNode\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 middleNode(self, head: ListNode) -> 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* middleNode(struct ListNode* head){\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 MiddleNode(ListNode head) {\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 * @return {ListNode}\n */\nvar middleNode = function(head) {\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# @return {ListNode}\ndef middle_node(head)\n\nend",
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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 middleNode(head: ListNode?): 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 middle_node(head: Option<Box<ListNode>>) -> 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 (middle-node head)\n  (-> (or/c list-node? #f) (or/c list-node? #f))\n\n  )",
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