leetcode-problemset/leetcode-cn/originData/reorder-list.json

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        "question": {
            "questionId": "143",
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            "categoryTitle": "Algorithms",
            "boundTopicId": 1303,
            "title": "Reorder List",
            "titleSlug": "reorder-list",
            "content": "<p>You are given the head of a singly linked-list. The list can be represented as:</p>\n\n<pre>\nL<sub>0</sub> &rarr; L<sub>1</sub> &rarr; &hellip; &rarr; L<sub>n - 1</sub> &rarr; L<sub>n</sub>\n</pre>\n\n<p><em>Reorder the list to be on the following form:</em></p>\n\n<pre>\nL<sub>0</sub> &rarr; L<sub>n</sub> &rarr; L<sub>1</sub> &rarr; L<sub>n - 1</sub> &rarr; L<sub>2</sub> &rarr; L<sub>n - 2</sub> &rarr; &hellip;\n</pre>\n\n<p>You may not modify 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/2021/03/04/reorder1linked-list.jpg\" style=\"width: 422px; height: 222px;\" />\n<pre>\n<strong>Input:</strong> head = [1,2,3,4]\n<strong>Output:</strong> [1,4,2,3]\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/03/09/reorder2-linked-list.jpg\" style=\"width: 542px; height: 222px;\" />\n<pre>\n<strong>Input:</strong> head = [1,2,3,4,5]\n<strong>Output:</strong> [1,5,2,4,3]\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, 5 * 10<sup>4</sup>]</code>.</li>\n\t<li><code>1 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n",
            "translatedTitle": "重排链表",
            "translatedContent": "<p>给定一个单链表 <code>L</code><em> </em>的头节点 <code>head</code> ，单链表 <code>L</code> 表示为：</p>\n\n<pre>\nL<sub>0</sub> → L<sub>1</sub> → … → L<sub>n - 1</sub> → L<sub>n</sub>\n</pre>\n\n<p>请将其重新排列后变为：</p>\n\n<pre>\nL<sub>0</sub> → L<sub>n</sub> → L<sub>1</sub> → L<sub>n - 1</sub> → L<sub>2</sub> → L<sub>n - 2</sub> → …</pre>\n\n<p>不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://pic.leetcode-cn.com/1626420311-PkUiGI-image.png\" style=\"width: 240px; \" /></p>\n\n<pre>\n<strong>输入：</strong>head = [1,2,3,4]\n<strong>输出：</strong>[1,4,2,3]</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://pic.leetcode-cn.com/1626420320-YUiulT-image.png\" style=\"width: 320px; \" /></p>\n\n<pre>\n<strong>输入：</strong>head = [1,2,3,4,5]\n<strong>输出：</strong>[1,5,2,4,3]</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>链表的长度范围为 <code>[1, 5 * 10<sup>4</sup>]</code></li>\n\t<li><code>1 &lt;= node.val &lt;= 1000</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    void reorderList(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 void reorderList(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 reorderList(self, head):\n        \"\"\"\n        :type head: ListNode\n        :rtype: None Do not return anything, modify head in-place instead.\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 reorderList(self, head: Optional[ListNode]) -> None:\n        \"\"\"\n        Do not return anything, modify head in-place instead.\n        \"\"\"",
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                    "code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n *     int val;\n *     struct ListNode *next;\n * };\n */\nvoid reorderList(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 void ReorderList(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 {void} Do not return anything, modify head in-place instead.\n */\nvar reorderList = function(head) {\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\n/**\n Do not return anything, modify head in-place instead.\n */\nfunction reorderList(head: ListNode | null): void {\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 reorderList(head: ListNode?): Unit {\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  void reorderList(ListNode? head) {\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 reorderList(head *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# @return {Void} Do not return anything, modify head in-place instead.\ndef reorder_list(head)\n\nend",
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