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        "question": {
            "questionId": "100199",
            "questionFrontendId": "面试题 08.05",
            "categoryTitle": "LCCI",
            "boundTopicId": 49281,
            "title": "Recursive Mulitply LCCI",
            "titleSlug": "recursive-mulitply-lcci",
            "content": "<p>Write a recursive function to multiply two positive integers without using the * operator. You can use addition, subtraction, and bit shifting, but you should minimize the number of those operations.</p>\r\n\r\n<p><strong>Example 1:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: A = 1, B = 10\r\n<strong> Output</strong>: 10\r\n</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: A = 3, B = 4\r\n<strong> Output</strong>: 12\r\n</pre>\r\n\r\n<p><strong>Note:</strong></p>\r\n\r\n<ol>\r\n\t<li>The result will not overflow.</li>\r\n</ol>\r\n",
            "translatedTitle": "递归乘法",
            "translatedContent": "<p>递归乘法。 写一个递归函数,不使用 * 运算符, 实现两个正整数的相乘。可以使用加号、减号、位移,但要吝啬一些。</p>\n\n<p> <strong>示例1:</strong></p>\n\n<pre>\n<strong> 输入</strong>:A = 1, B = 10\n<strong> 输出</strong>:10\n</pre>\n\n<p> <strong>示例2:</strong></p>\n\n<pre>\n<strong> 输入</strong>:A = 3, B = 4\n<strong> 输出</strong>:12\n</pre>\n\n<p> <strong>提示:</strong></p>\n\n<ol>\n<li>保证乘法范围不会溢出</li>\n</ol>\n",
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            "topicTags": [
                {
                    "name": "Bit Manipulation",
                    "slug": "bit-manipulation",
                    "translatedName": "位运算",
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                    "name": "Recursion",
                    "slug": "recursion",
                    "translatedName": "递归",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    int multiply(int A, int B) {\n\n    }\n};",
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                    "code": "class Solution {\n    public int multiply(int A, int B) {\n\n    }\n}",
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                    "langSlug": "python",
                    "code": "class Solution(object):\n    def multiply(self, A, B):\n        \"\"\"\n        :type A: int\n        :type B: int\n        :rtype: int\n        \"\"\"",
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                    "code": "\n\nint multiply(int A, int B){\n\n}\n",
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                    "code": "public class Solution {\n    public int Multiply(int A, int B) {\n\n    }\n}",
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                    "code": "/**\n * @param {number} A\n * @param {number} B\n * @return {number}\n */\nvar multiply = function(A, B) {\n\n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $A\n     * @param Integer $B\n     * @return Integer\n     */\n    function multiply($A, $B) {\n\n    }\n}",
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                    "code": "class Solution {\n    func multiply(_ A: Int, _ B: Int) -> Int {\n\n    }\n}",
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                    "code": "class Solution {\n    fun multiply(A: Int, B: Int): Int {\n\n    }\n}",
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                    "code": "func multiply(A int, B int) int {\n\n}",
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                    "code": "# @param {Integer} a\n# @param {Integer} b\n# @return {Integer}\ndef multiply(a, b)\n\nend",
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                    "code": "defmodule Solution do\n  @spec multiply(a :: integer, b :: integer) :: integer\n  def multiply(a, b) do\n\n  end\nend",
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            "hints": [
                "考虑将8乘以9看作是计算宽度为8、高度为9的矩阵中的单元数。",
                "如果你想计算8×9矩阵中的单元格数,可以先计算4×9矩阵中的单元格数,然后加倍。",
                "想想你如何处理奇数。",
                "如果不同的递归调用有重复的工作,你可以缓存它吗?",
                "如果你在做9×7(都是奇数),那么你可以换成4×7和5×7。",
                "或者,如果你在计算9×7,可以计算4×7,加倍,然后再加7。"
            ],
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