leetcode-problemset/leetcode-cn/originData/closed-number-lcci.json

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        "question": {
            "questionId": "100183",
            "questionFrontendId": "面试题 05.04",
            "categoryTitle": "LCCI",
            "boundTopicId": 46357,
            "title": "Closed Number LCCI",
            "titleSlug": "closed-number-lcci",
            "content": "<p>Given a positive integer, print the next smallest and the next largest number that have the same number of 1 bits in their binary representation.</p>\r\n\r\n<p><strong>Example1:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: num = 2 (0b10)\r\n<strong> Output</strong>: [4, 1] ([0b100, 0b1])\r\n</pre>\r\n\r\n<p><strong>Example2:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: num = 1\r\n<strong> Output</strong>: [2, -1]\r\n</pre>\r\n\r\n<p><strong>Note:</strong></p>\r\n\r\n<ol>\r\n\t<li><code>1 &lt;= num &lt;=&nbsp;2147483647</code></li>\r\n\t<li>If there is no next smallest or next largest number, output -1.</li>\r\n</ol>\r\n",
            "translatedTitle": "下一个数",
            "translatedContent": "<p>下一个数。给定一个正整数，找出与其二进制表达式中1的个数相同且大小最接近的那两个数（一个略大，一个略小）。</p>\n\n<p> <strong>示例1:</strong></p>\n\n<pre>\n<strong> 输入</strong>：num = 2（或者0b10）\n<strong> 输出</strong>：[4, 1] 或者（[0b100, 0b1]）\n</pre>\n\n<p> <strong>示例2:</strong></p>\n\n<pre>\n<strong> 输入</strong>：num = 1\n<strong> 输出</strong>：[2, -1]\n</pre>\n\n<p> <strong>提示:</strong></p>\n\n<ol>\n<li><code>num</code>的范围在[1, 2147483647]之间；</li>\n<li>如果找不到前一个或者后一个满足条件的正数，那么输出 -1。</li>\n</ol>\n",
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                {
                    "name": "Bit Manipulation",
                    "slug": "bit-manipulation",
                    "translatedName": "位运算",
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                    "code": "class Solution {\npublic:\n    vector<int> findClosedNumbers(int num) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def findClosedNumbers(self, num):\n        \"\"\"\n        :type num: 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* findClosedNumbers(int num, int* returnSize){\n\n}\n",
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                    "code": "func findClosedNumbers(num int) []int {\n\n}",
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            "hints": [
                "下一步：从每个蛮力解法开始。",
                "下一个：想象一个二进制数，在整个数中分布一串1和0。假设你把一个1翻转成0，把一个0翻转成1。在什么情况下数会更大？在什么情况下数会更小？",
                "下一步：如果你将1翻转成0，0翻转成1，假设 0 -> 1位更大，那么它就会变大。你如何使用这个来创建下一个最大的数字（具有相同数量的1）？",
                "下一步：你能翻转0到1，创建下一个最大的数字吗？",
                "下一步：把0翻转为1将创建一个更大的数字。索引越靠右，数字越大。如果有一个1001这样的数字，那么我们就想翻转最右边的0（创建1011）。但是如果有一个1010这样的数字，我们就不应该翻转最右边的1。",
                "下一步：我们应该翻转最右边但非拖尾的0。数字1010会变成1110。完成后，我们需要把1翻转成0让数字尽可能小，但要大于原始数字（1010）。该怎么办？如何缩小数字？",
                "下一步：我们可以通过将所有的1移动到翻转位的右侧，并尽可能地向右移动来缩小数字（在这个过程中去掉一个1）。",
                "获取前一个：一旦你解决了“获取后一个”，请尝试翻转“获取前一个”的逻辑。"
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