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{
"data": {
"question": {
"questionId": "1000260",
"questionFrontendId": "剑指 Offer II 024",
"categoryTitle": "LCOF2",
"boundTopicId": 910263,
"title": "反转链表",
"titleSlug": "UHnkqh",
"content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
"translatedTitle": "反转链表",
"translatedContent": "<p>给定单链表的头节点 <code>head</code> ,请反转链表,并返回反转后的链表的头节点。</p>\n\n<div class=\"original__bRMd\">\n<div>\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg\" style=\"width: 302px; \" />\n<pre>\n<strong>输入:</strong>head = [1,2,3,4,5]\n<strong>输出:</strong>[5,4,3,2,1]\n</pre>\n\n<p><strong>示例 2</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg\" style=\"width: 102px;\" />\n<pre>\n<strong>输入:</strong>head = [1,2]\n<strong>输出:</strong>[2,1]\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>head = []\n<strong>输出:</strong>[]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>链表中节点的数目范围是 <code>[0, 5000]</code></li>\n\t<li><code>-5000 &lt;= Node.val &lt;= 5000</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶:</strong>链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题?</p>\n</div>\n</div>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意:本题与主站 206&nbsp;题相同:&nbsp;<a href=\"https://leetcode-cn.com/problems/reverse-linked-list/\">https://leetcode-cn.com/problems/reverse-linked-list/</a></p>\n",
"isPaidOnly": false,
"difficulty": "Easy",
"likes": 60,
"dislikes": 0,
"isLiked": null,
"similarQuestions": "[]",
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{
"username": "LeetCode",
"profileUrl": "/u/leetcode",
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"name": "Recursion",
"slug": "recursion",
"translatedName": "递归",
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"name": "Linked List",
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"translatedName": "链表",
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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* reverseList(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 reverseList(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 reverseList(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 reverseList(self, head: ListNode) -> ListNode:",
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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 reverseList = 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 reverse_list(head)\n\nend",
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"code": "/**\n * Definition for singly-linked list.\n * type ListNode struct {\n * Val int\n * Next *ListNode\n * }\n */\nfunc reverseList(head *ListNode) *ListNode {\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 reverseList(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 reverse_list(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 (reverse-list head)\n (-> (or/c list-node? #f) (or/c list-node? #f))\n\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 reverse_list(head :: ListNode.t | nil) :: ListNode.t | nil\n def reverse_list(head) do\n\n end\nend",
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