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leetcode-problemset/leetcode-cn/originData/partition-list-lcci.json
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{
"data": {
"question": {
"questionId": "1000008",
"questionFrontendId": "面试题 02.04",
"categoryTitle": "LCCI",
"boundTopicId": 91433,
"title": "Partition List LCCI",
"titleSlug": "partition-list-lcci",
"content": "<p>Write code to partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x. If x is contained within the list, the values of x only need to be after the elements less than x (see below). The partition element x can appear anywhere in the &quot;right partition&quot;; it does not need to appear between the left and right partitions.</p>\r\n\r\n<p><strong>Example:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> head = 3-&gt;5-&gt;8-&gt;5-&gt;10-&gt;2-&gt;1, <em>x</em> = 5\r\n<strong>Output:</strong> 3-&gt;1-&gt;2-&gt;10-&gt;5-&gt;5-&gt;8\r\n</pre>\r\n",
"translatedTitle": "分割链表",
"translatedContent": "<p>给你一个链表的头节点 <code>head</code> 和一个特定值<em> </em><code>x</code> ,请你对链表进行分隔,使得所有 <strong>小于</strong> <code>x</code> 的节点都出现在 <strong>大于或等于</strong> <code>x</code> 的节点之前。</p>\n\n<p>你不需要&nbsp;<strong>保留</strong>&nbsp;每个分区中各节点的初始相对位置。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/04/partition.jpg\" style=\"width: 662px; height: 222px;\" />\n<pre>\n<strong>输入:</strong>head = [1,4,3,2,5,2], x = 3\n<strong>输出</strong>[1,2,2,4,3,5]\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>head = [2,1], x = 2\n<strong>输出</strong>[1,2]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>链表中节点的数目在范围 <code>[0, 200]</code> 内</li>\n\t<li><code>-100 &lt;= Node.val &lt;= 100</code></li>\n\t<li><code>-200 &lt;= x &lt;= 200</code></li>\n</ul>\n",
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"difficulty": "Medium",
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"code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode(int x) : val(x), next(NULL) {}\n * };\n */\nclass Solution {\npublic:\n ListNode* partition(ListNode* head, int x) {\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(int x) { val = x; }\n * }\n */\nclass Solution {\n public ListNode partition(ListNode head, int x) {\n\n }\n}",
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"code": "# Definition for singly-linked list.\n# class ListNode(object):\n# def __init__(self, x):\n# self.val = x\n# self.next = None\n\nclass Solution(object):\n def partition(self, head, x):\n \"\"\"\n :type head: ListNode\n :type x: int\n :rtype: ListNode\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 x) { val = x; }\n * }\n */\npublic class Solution {\n public ListNode Partition(ListNode head, int x) {\n\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 partition(head: ListNode | null, x: number): ListNode | null {\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(_ val: Int) {\n * self.val = val\n * self.next = nil\n * }\n * }\n */\nclass Solution {\n func partition(_ head: ListNode?, _ x: 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 partition(head *ListNode, x 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)\n# @val = val\n# @next = nil\n# end\n# end\n\n# @param {ListNode} head\n# @param {Integer} x\n# @return {ListNode}\ndef partition(head, x)\n\nend",
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"code": "/**\n * Definition for singly-linked list.\n * class ListNode(var _x: Int = 0) {\n * var next: ListNode = null\n * var x: Int = _x\n * }\n */\nobject Solution {\n def partition(head: ListNode, x: 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 partition(head: Option<Box<ListNode>>, x: i32) -> Option<Box<ListNode>> {\n\n }\n}",
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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 partition(Head :: #list_node{} | null, X :: integer()) -> #list_node{} | null.\npartition(Head, X) ->\n .",
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"code": "# Definition for singly-linked list.\n#\n# defmodule ListNode do\n# @type t :: %__MODULE__{\n# val: integer,\n# next: ListNode.t() | nil\n# }\n# defstruct val: 0, next: nil\n# end\n\ndefmodule Solution do\n @spec partition(head :: ListNode.t | nil, x :: integer) :: ListNode.t | nil\n def partition(head, x) do\n\n end\nend",
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"hints": [
"这个问题有很多解法,其中大部分都有最优的运行时间。有些代码比其他代码更短,更干净。你可以想出不同的解法吗?",
"考虑元素不必保持相同的相对顺序。我们只需要确保小于基准点的元素必须位于比基准点大的元素之前。这有助于你想出更多的解法吗?"
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