mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
192 lines
19 KiB
JSON
192 lines
19 KiB
JSON
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"questionFrontendId": "面试题 08.05",
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"categoryTitle": "LCCI",
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"title": "Recursive Mulitply LCCI",
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"titleSlug": "recursive-mulitply-lcci",
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"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",
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"translatedTitle": "递归乘法",
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"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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"difficulty": "Medium",
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"name": "Bit Manipulation",
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"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": "# @param {Integer} a\n# @param {Integer} b\n# @return {Integer}\ndef multiply(a, b)\n\nend",
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"code": "-spec multiply(A :: integer(), B :: integer()) -> integer().\nmultiply(A, B) ->\n .",
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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": [
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"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。",
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"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。",
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"想想你如何处理奇数。",
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"如果不同的递归调用有重复的工作，你可以缓存它吗？",
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"如果你在做9×7（都是奇数），那么你可以换成4×7和5×7。",
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"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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