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leetcode-problemset/leetcode-cn/originData/missing-number-lcci.json
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{
"data": {
"question": {
"questionId": "1000032",
"questionFrontendId": "面试题 17.04",
"categoryTitle": "LCCI",
"boundTopicId": 94394,
"title": "Missing Number LCCI",
"titleSlug": "missing-number-lcci",
"content": "<p>An array&nbsp;contains all the integers from 0 to n, except for one number which is missing.&nbsp; Write code to find the missing integer. Can you do it in O(n) time?</p>\r\n\r\n<p><strong>Note: </strong>This problem is slightly different from the original one the book.</p>\r\n\r\n<p><strong>Example 1: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>[3,0,1]\r\n<strong>Output: </strong>2</pre>\r\n\r\n<p>&nbsp;</p>\r\n\r\n<p><strong>Example 2: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>[9,6,4,2,3,5,7,0,1]\r\n<strong>Output: </strong>8\r\n</pre>\r\n",
"translatedTitle": "消失的数字",
"translatedContent": "<p>数组<code>nums</code>包含从<code>0</code>到<code>n</code>的所有整数但其中缺了一个。请编写代码找出那个缺失的整数。你有办法在O(n)时间内完成吗?</p>\n\n<p><strong>注意:</strong>本题相对书上原题稍作改动</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre><strong>输入:</strong>[3,0,1]\n<strong>输出:</strong>2</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 2</strong></p>\n\n<pre><strong>输入:</strong>[9,6,4,2,3,5,7,0,1]\n<strong>输出:</strong>8\n</pre>\n",
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"difficulty": "Easy",
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"name": "Bit Manipulation",
"slug": "bit-manipulation",
"translatedName": "位运算",
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"translatedName": "数组",
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"lang": "C++",
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"code": "class Solution {\npublic:\n int missingNumber(vector<int>& nums) {\n\n }\n};",
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"code": "class Solution {\n public int missingNumber(int[] nums) {\n\n }\n}",
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"lang": "Python",
"langSlug": "python",
"code": "class Solution(object):\n def missingNumber(self, nums):\n \"\"\"\n :type nums: List[int]\n :rtype: int\n \"\"\"",
"__typename": "CodeSnippetNode"
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"code": "class Solution:\n def missingNumber(self, nums: List[int]) -> int:",
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"lang": "C",
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"code": "\n\nint missingNumber(int* nums, int numsSize){\n\n}\n",
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"code": "public class Solution {\n public int MissingNumber(int[] nums) {\n\n }\n}",
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"lang": "JavaScript",
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"code": "/**\n * @param {number[]} nums\n * @return {number}\n */\nvar missingNumber = function(nums) {\n\n};",
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"code": "function missingNumber(nums: number[]): number {\n\n};",
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"code": "class Solution {\n\n /**\n * @param Integer[] $nums\n * @return Integer\n */\n function missingNumber($nums) {\n\n }\n}",
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"lang": "Swift",
"langSlug": "swift",
"code": "class Solution {\n func missingNumber(_ nums: [Int]) -> Int {\n\n }\n}",
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"lang": "Kotlin",
"langSlug": "kotlin",
"code": "class Solution {\n fun missingNumber(nums: IntArray): Int {\n\n }\n}",
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"code": "class Solution {\n int missingNumber(List<int> nums) {\n\n }\n}",
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"code": "impl Solution {\n pub fn missing_number(nums: Vec<i32>) -> i32 {\n\n }\n}",
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{
"lang": "Erlang",
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},
{
"lang": "Elixir",
"langSlug": "elixir",
"code": "defmodule Solution do\n @spec missing_number(nums :: [integer]) :: integer\n def missing_number(nums) do\n\n end\nend",
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"hints": [
"你需要多长时间才能算出缺失数字的最小有效位?",
"要找到缺失的数字中的最小有效位,你其实知道有多少个 0 和 1。例如如果你看到最小有效位有 3 个 0 和 3 个 1那么缺失的数字的最小值必定是 1。想想看:在任何 0 和 1 的序列中,你会得到 0然后是 1然后又是 0然后又是 1以此类推。",
"一旦确定最小有效位是 0(或 1),就可以排除所有不以 0 作为最小有效位的数。这个问题和前面的有什么不同?"
],
"solution": null,
"status": null,
"sampleTestCase": "[3,0,1]",
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