mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
206 lines
20 KiB
JSON
206 lines
20 KiB
JSON
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"questionId": "1000021",
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"questionFrontendId": "面试题 17.14",
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"categoryTitle": "LCCI",
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"boundTopicId": 93044,
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"title": "Smallest K LCCI",
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"content": "<p>Design an algorithm to find the smallest K numbers in an array.</p>\r\n\r\n<p><strong>Example: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong> arr = [1,3,5,7,2,4,6,8], k = 4\r\n<strong>Output: </strong> [1,2,3,4]\r\n</pre>\r\n\r\n<p><strong>Note: </strong></p>\r\n\r\n<ul>\r\n\t<li><code>0 <= len(arr) <= 100000</code></li>\r\n\t<li><code>0 <= k <= min(100000, len(arr))</code></li>\r\n</ul>\r\n",
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"translatedTitle": "最小K个数",
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"translatedContent": "<p>设计一个算法，找出数组中最小的k个数。以任意顺序返回这k个数均可。</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong> arr = [1,3,5,7,2,4,6,8], k = 4\n<strong>输出：</strong> [1,2,3,4]\n</pre>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= len(arr) <= 100000</code></li>\n\t<li><code>0 <= k <= min(100000, len(arr))</code></li>\n</ul>\n",
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"difficulty": "Medium",
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"name": "Divide and Conquer",
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"name": "Quickselect",
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"slug": "quickselect",
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"translatedName": "快速选择",
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"name": "Heap (Priority Queue)",
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"code": "class Solution {\npublic:\n vector<int> smallestK(vector<int>& arr, int k) {\n\n }\n};",
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"code": "class Solution(object):\n def smallestK(self, arr, k):\n \"\"\"\n :type arr: List[int]\n :type k: int\n :rtype: List[int]\n \"\"\"",
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"code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* smallestK(int* arr, int arrSize, int k, int* returnSize){\n\n}\n",
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"code": "# @param {Integer[]} arr\n# @param {Integer} k\n# @return {Integer[]}\ndef smallest_k(arr, k)\n\nend",
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"code": "impl Solution {\n pub fn smallest_k(arr: Vec<i32>, k: i32) -> Vec<i32> {\n\n }\n}",
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"code": "(define/contract (smallest-k arr k)\n (-> (listof exact-integer?) exact-integer? (listof exact-integer?))\n\n )",
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"code": "-spec smallest_k(Arr :: [integer()], K :: integer()) -> [integer()].\nsmallest_k(Arr, K) ->\n .",
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"code": "defmodule Solution do\n @spec smallest_k(arr :: [integer], k :: integer) :: [integer]\n def smallest_k(arr, k) do\n\n end\nend",
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"hints": [
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"实际上有几种方法。动脑筋想一想。从简单的方法开始也没问题。",
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"考虑以某种方式重新组织数据或者使用其他数据结构。",
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"你能把这些数字排序吗?",
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"使用堆或某种树怎么样?",
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"如果你选了一个任意的元素，那么需要多长时间才能算出它的元素的排序（比它大或比它小的元素的个数）?",
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"如果你选择一个任意的元素，平均来说，就会得到一个在第50百分位数附近的元素（一半的元素比它大，一半的元素比它小）。如果反复这样做呢?",
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"回想一下前面的提示，特别是与快速排序相关的提示。",
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"如果当你选择一个元素时，你交换周围的元素（就像在快速排序中所做的那样），使它所有下方的元素都位于上方的元素之前，那会怎么样？如果你重复做这个，能找到最小的一百万个数吗？"
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],
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