leetcode-problemset/leetcode-cn/originData/insert-into-bits-lcci.json

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        "question": {
            "questionId": "100180",
            "questionFrontendId": "面试题 05.01",
            "categoryTitle": "LCCI",
            "boundTopicId": 46259,
            "title": "Insert Into Bits LCCI",
            "titleSlug": "insert-into-bits-lcci",
            "content": "<p>You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to insert M into N such that M starts at bit j and ends at bit i. You can assume that the bits j through i have enough space to fit all of M. That is, if M = 10011, you can assume that there are at least 5 bits between j and i. You would not, for example, have j = 3 and i = 2, because M could not fully fit between bit 3 and bit 2.</p>\r\n\r\n<p><strong>Example1:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: N = 10000000000, M = 10011, i = 2, j = 6\r\n<strong> Output</strong>: N = 10001001100\r\n</pre>\r\n\r\n<p><strong>Example2:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>:  N = 0, M = 11111, i = 0, j = 4\r\n<strong> Output</strong>: N = 11111\r\n</pre>\r\n",
            "translatedTitle": "插入",
            "translatedContent": "<p>给定两个整型数字 <code>N</code> 与 <code>M</code>，以及表示比特位置的 <code>i</code> 与 <code>j</code>（<code>i <= j</code>，且从 0 位开始计算）。</p>\n\n<p>编写一种方法，使 <code>M</code> 对应的二进制数字插入 <code>N</code> 对应的二进制数字的第 <code>i ~ j</code> 位区域，不足之处用 <code>0</code> 补齐。具体插入过程如图所示。</p>\n\n<p><img alt=\"\" src=\"https://pic.leetcode-cn.com/1610104070-NuLVQi-05.01.gif\" style=\"width: 267px; height: 200px;\" /></p>\n\n<p>题目保证从 <code>i</code> 位到 <code>j</code> 位足以容纳 <code>M</code>， 例如： <code>M = 10011</code>，则 <code>i～j</code> 区域至少可容纳 5 位。</p>\n\n<p> </p>\n\n<p><strong>示例1:</strong></p>\n\n<pre>\n<strong> 输入</strong>：N = 1024(10000000000), M = 19(10011), i = 2, j = 6\n<strong> 输出</strong>：N = 1100(10001001100)\n</pre>\n\n<p><strong>示例2:</strong></p>\n\n<pre>\n<strong> 输入</strong>： N = 0, M = 31(11111), i = 0, j = 4\n<strong> 输出</strong>：N = 31(11111)\n</pre>\n",
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                {
                    "name": "Bit Manipulation",
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                    "code": "class Solution(object):\n    def insertBits(self, N, M, i, j):\n        \"\"\"\n        :type N: int\n        :type M: int\n        :type i: int\n        :type j: int\n        :rtype: int\n        \"\"\"",
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                    "code": "/**\n * @param {number} N\n * @param {number} M\n * @param {number} i\n * @param {number} j\n * @return {number}\n */\nvar insertBits = function(N, M, i, j) {\n\n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $N\n     * @param Integer $M\n     * @param Integer $i\n     * @param Integer $j\n     * @return Integer\n     */\n    function insertBits($N, $M, $i, $j) {\n\n    }\n}",
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                    "code": "# @param {Integer} n\n# @param {Integer} m\n# @param {Integer} i\n# @param {Integer} j\n# @return {Integer}\ndef insert_bits(n, m, i, j)\n\nend",
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            "hints": [
                "把这个分成几个部分。先将精力放在清除适当的位上。",
                "要清除这些位，创建一个看起来像是一系列1，然后是0，然后是1的“位掩码”。",
                "在开始或结束时很容易创建一个0的位掩码。但是，有一堆0时，你如何在中间创建一个零位掩码？简单的做法是，为左侧创建一个位掩码，然后为右侧创建一个位掩码。然后你可以合并两边。"
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