mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
185 lines
23 KiB
JSON
185 lines
23 KiB
JSON
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"questionId": "1000007",
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"questionFrontendId": "面试题 02.02",
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"categoryTitle": "LCCI",
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"boundTopicId": 91405,
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"title": "Kth Node From End of List LCCI",
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"titleSlug": "kth-node-from-end-of-list-lcci",
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"content": "<p>Implement an algorithm to find the kth to last element of a singly linked list. Return the value of the element.</p>\r\n\r\n<p><strong>Note: </strong>This problem is slightly different from the original one in the book.</p>\r\n\r\n<p><strong>Example: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong> 1->2->3->4->5 和 <em>k</em> = 2\r\n<strong>Output: </strong>4</pre>\r\n\r\n<p><strong>Note: </strong></p>\r\n\r\n<p>k is always valid.</p>\r\n",
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"translatedTitle": "返回倒数第 k 个节点",
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"translatedContent": "<p>实现一种算法，找出单向链表中倒数第 k 个节点。返回该节点的值。</p>\n\n<p><strong>注意：</strong>本题相对原题稍作改动</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong> 1->2->3->4->5 和 <em>k</em> = 2\n<strong>输出： </strong>4</pre>\n\n<p><strong>说明：</strong></p>\n\n<p>给定的 <em>k</em> 保证是有效的。</p>\n",
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"difficulty": "Easy",
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{
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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(int x) : val(x), next(NULL) {}\n * };\n */\nclass Solution {\npublic:\n int kthToLast(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(int x) { val = x; }\n * }\n */\nclass Solution {\n public int kthToLast(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, x):\n# self.val = x\n# self.next = None\n\nclass Solution(object):\n def kthToLast(self, head, k):\n \"\"\"\n :type head: ListNode\n :type k: int\n :rtype: int\n \"\"\"",
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"code": "# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, x):\n# self.val = x\n# self.next = None\n\nclass Solution:\n def kthToLast(self, head: ListNode, k: int) -> int:",
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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\nint kthToLast(struct ListNode* head, int k){\n\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 x) { val = x; }\n * }\n */\npublic class Solution {\n public int KthToLast(ListNode head, int k) {\n\n }\n}",
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"code": "/**\n * Definition for singly-linked list.\n * function ListNode(val) {\n * this.val = val;\n * this.next = null;\n * }\n */\n/**\n * @param {ListNode} head\n * @param {number} k\n * @return {number}\n */\nvar kthToLast = function(head, k) {\n\n};",
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"lang": "Ruby",
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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} k\n# @return {Integer}\ndef kth_to_last(head, k)\n\nend",
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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 kthToLast(_ head: ListNode?, _ k: Int) -> Int {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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{
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"lang": "Go",
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"code": "/**\n * Definition for singly-linked list.\n * type ListNode struct {\n * Val int\n * Next *ListNode\n * }\n */\nfunc kthToLast(head *ListNode, k int) int {\n\n}",
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"lang": "Scala",
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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 kthToLast(head: ListNode, k: Int): Int = {\n\n }\n}",
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"lang": "Kotlin",
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"langSlug": "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 kthToLast(head: ListNode?, k: Int): Int {\n\n }\n}",
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"lang": "Rust",
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"langSlug": "rust",
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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 kth_to_last(head: Option<Box<ListNode>>, k: i32) -> i32 {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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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) { $this->val = $val; }\n * }\n */\nclass Solution {\n\n /**\n * @param ListNode $head\n * @param Integer $k\n * @return Integer\n */\n function kthToLast($head, $k) {\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 kthToLast(head: ListNode | null, k: number): number {\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 (kth-to-last head k)\n (-> (or/c list-node? #f) exact-integer? exact-integer?)\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 kth_to_last(Head :: #list_node{} | null, K :: integer()) -> integer().\nkth_to_last(Head, K) ->\n .",
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"lang": "Elixir",
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"langSlug": "elixir",
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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 kth_to_last(head :: ListNode.t | nil, k :: integer) :: integer\n def kth_to_last(head, k) do\n\n end\nend",
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"hints": [
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"如果你知道链表大小，会怎么样？找到最后第k个元素和找到第x个元素有何区别？",
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"如果你不知道链表的大小，你能计算它吗？这将如何影响运行时间？",
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"尝试用递归法实现。如果你能找到(k − 1)到最后一个元素，可以找到第k个元素吗？",
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"你可能会发现返回多个值大有用处。有些语言不直接支持这一点，但基本上使用任何语言都有解决方法。这些解决方法有哪些？",
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"你能通过递归做到吗？想象一下，如果有两个指针指向相邻节点，它们通过链表以相同的速度移动。当一个到达链表的结尾时，另一个在哪里？"
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