leetcode-problemset/leetcode-cn/originData/memoize.json

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        "question": {
            "questionId": "2731",
            "questionFrontendId": "2623",
            "categoryTitle": "JavaScript",
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            "title": "Memoize",
            "titleSlug": "memoize",
            "content": "<p>Given a function <code>fn</code>, return a&nbsp;<strong>memoized</strong>&nbsp;version of that function.</p>\n\n<p>A&nbsp;<strong>memoized&nbsp;</strong>function is a function that will never be called twice with&nbsp;the same inputs. Instead it will returned a cached value.</p>\n\n<p>You can assume there are&nbsp;<strong>3&nbsp;</strong>possible input functions:&nbsp;<code>sum</code><strong>, </strong><code>fib</code><strong>,&nbsp;</strong>and&nbsp;<code>factorial</code><strong>.</strong></p>\n\n<ul>\n\t<li><code>sum</code><strong>&nbsp;</strong>accepts two integers&nbsp;<code>a</code> and <code>b</code> and returns <code>a + b</code>.</li>\n\t<li><code>fib</code><strong>&nbsp;</strong>accepts a&nbsp;single integer&nbsp;<code>n</code> and&nbsp;returns&nbsp;<code>1</code> if <font face=\"monospace\"><code>n &lt;= 1</code> </font>or<font face=\"monospace\">&nbsp;<code>fib(n - 1) + fib(n - 2)</code>&nbsp;</font>otherwise.</li>\n\t<li><code>factorial</code>&nbsp;accepts a single integer&nbsp;<code>n</code> and returns <code>1</code>&nbsp;if&nbsp;<code>n &lt;= 1</code>&nbsp;or&nbsp;<code>factorial(n - 1) * n</code>&nbsp;otherwise.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n&quot;sum&quot;\n[&quot;call&quot;,&quot;call&quot;,&quot;getCallCount&quot;,&quot;call&quot;,&quot;getCallCount&quot;]\n[[2,2],[2,2],[],[1,2],[]]\n<strong>Output</strong>\n[4,4,1,3,2]\n\n<strong>Explanation</strong>\nconst sum = (a, b) =&gt; a + b;\nconst memoizedSum = memoize(sum);\nmemoizedSum(2, 2); // Returns 4. sum() was called as (2, 2) was not seen before.\nmemoizedSum(2, 2); // Returns 4. However sum() was not called because the same inputs were seen before.\n// Total call count: 1\nmemoizedSum(1, 2); // Returns 3. sum() was called as (1, 2) was not seen before.\n// Total call count: 2\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input\n</strong>&quot;factorial&quot;\n[&quot;call&quot;,&quot;call&quot;,&quot;call&quot;,&quot;getCallCount&quot;,&quot;call&quot;,&quot;getCallCount&quot;]\n[[2],[3],[2],[],[3],[]]\n<strong>Output</strong>\n[2,6,2,2,6,2]\n\n<strong>Explanation</strong>\nconst factorial = (n) =&gt; (n &lt;= 1) ? 1 : (n * factorial(n - 1));\nconst memoFactorial = memoize(factorial);\nmemoFactorial(2); // Returns 2.\nmemoFactorial(3); // Returns 6.\nmemoFactorial(2); // Returns 2. However factorial was not called because 2 was seen before.\n// Total call count: 2\nmemoFactorial(3); // Returns 6. However factorial was not called because 3 was seen before.\n// Total call count: 2\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input\n</strong>&quot;fib&quot;\n[&quot;call&quot;,&quot;getCallCount&quot;]\n[[5],[]]\n<strong>Output</strong>\n[8,1]\n\n<strong>Explanation\n</strong>fib(5) = 8\n// Total call count: 1\n\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= a, b &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= n &lt;= 10</code></li>\n\t<li><code>at most 10<sup>5</sup>&nbsp;function calls</code></li>\n\t<li><code>at most 10<sup>5</sup>&nbsp;attempts to access callCount</code></li>\n\t<li><code>input function is sum, fib, or factorial</code></li>\n</ul>\n",
            "translatedTitle": "记忆函数",
            "translatedContent": "<p>请你编写一个函数，它接收另一个函数作为输入，并返回该函数的 <strong>记忆化</strong> 后的结果。</p>\n\n<p><strong>记忆函数</strong> 是一个对于相同的输入永远不会被调用两次的函数。相反，它将返回一个缓存值。</p>\n\n<p>你可以假设有 <strong>3</strong> 个可能的输入函数：<code>sum</code> 、<code>fib</code> 和 <code>factorial</code> 。</p>\n\n<ul>\n\t<li>&nbsp;<code>sum</code> 接收两个整型参数 <code>a</code> 和 <code>b</code> ，并返回 <code>a + b</code> 。</li>\n\t<li>&nbsp;<code>fib</code> 接收一个整型参数&nbsp;<code>n</code> ，如果 <code>n &lt;= 1</code> 则返回 <code>1</code>，否则返回 <code>fib (n - 1) + fib (n - 2)</code>。</li>\n\t<li>&nbsp;<code>factorial</code> 接收一个整型参数 <code>n</code> ，如果 <code>n &lt;= 1</code> 则返回&nbsp;&nbsp;<code>1</code>&nbsp;，否则返回 <code>factorial(n - 1) * n</code> 。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\n\"sum\"\n[\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\n[[2,2],[2,2],[],[1,2],[]]\n<strong>输出：</strong>\n[4,4,1,3,2]\n\n<strong>解释：</strong>\nconst sum = (a, b) =&gt; a + b;\nconst memoizedSum = memoize(sum);\nmemoizedSum (2, 2);// 返回 4。sum() 被调用，因为之前没有使用参数 (2, 2) 调用过。\nmemoizedSum (2, 2);// 返回 4。没有调用 sum()，因为前面有相同的输入。\n//总调用数： 1\nmemoizedSum(1、2);// 返回 3。sum() 被调用，因为之前没有使用参数 (1, 2) 调用过。\n//总调用数： 2\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：\n</strong>\"factorial\"\n[\"call\",\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\n[[2],[3],[2],[],[3],[]]\n<strong>输出：</strong>\n[2,6,2,2,6,2]\n\n<strong>解释：</strong>\nconst factorial = (n) =&gt; (n &lt;= 1) ? 1 : (n * factorial(n - 1));\nconst memoFactorial = memoize(factorial);\nmemoFactorial(2); // 返回 2。\nmemoFactorial(3); // 返回 6。\nmemoFactorial(2); // 返回 2。 没有调用 factorial()，因为前面有相同的输入。\n// 总调用数：2\nmemoFactorial(3); // 返回 6。 没有调用 factorial()，因为前面有相同的输入。\n// 总调用数：2\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：\n</strong>\"fib\"\n[\"call\",\"getCallCount\"]\n[[5],[]]\n<strong>输出：</strong>\n[8,1]\n\n<strong>解释：\n</strong>fib(5) = 8\n// 总调用数：1\n\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= a, b &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= n &lt;= 10</code></li>\n\t<li><code>at most 10<sup>5</sup>&nbsp;function calls</code></li>\n\t<li><code>at most 10<sup>5</sup>&nbsp;attempts to access callCount</code></li>\n\t<li><code>input function is sum, fib, or factorial</code></li>\n</ul>\n",
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                    "code": "/**\n * @param {Function} fn\n */\nfunction memoize(fn) {\n    return function(...args) {\n        \n    }\n}\n\n\n/** \n * let callCount = 0;\n * const memoizedFn = memoize(function (a, b) {\n *\t callCount += 1;\n *   return a + b;\n * })\n * memoizedFn(2, 3) // 5\n * memoizedFn(2, 3) // 5\n * console.log(callCount) // 1 \n */",
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                    "code": "type Fn = (...params: any) => any\n\nfunction memoize(fn: Fn): Fn {\n    return function(...args) {\n        \n    }\n}\n\n\n/** \n * let callCount = 0;\n * const memoizedFn = memoize(function (a, b) {\n *\t callCount += 1;\n *   return a + b;\n * })\n * memoizedFn(2, 3) // 5\n * memoizedFn(2, 3) // 5\n * console.log(callCount) // 1 \n */",
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                "You can create copy of a function by spreading function parameters. \r\n\r\nfunction outerFunction(passedFunction) {\r\n  return newFunction(...params) {\r\n    return passedFunction(...params);\r\n  };\r\n}",
                "params is an array. Since you know all values in the array are numbers, you can turn it into a string with JSON.stringify().",
                "In the outerFunction, you can declare a Map or Object. In the inner function you can avoid executing the passed function if the params have already been passed before."
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            "exampleTestcases": "\"sum\"\n[\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\n[[2,2],[2,2],[],[1,2],[]]\n\"factorial\"\n[\"call\",\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\n[[2],[3],[2],[],[3],[]]\n\"fib\"\n[\"call\",\"getCallCount\"]\n[[5],[]]",
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