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https://gitee.com/coder-xiaomo/leetcode-problemset
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64 lines
11 KiB
JSON
64 lines
11 KiB
JSON
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"categoryTitle": "JavaScript",
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"boundTopicId": 2222279,
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"title": "Memoize II",
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"titleSlug": "memoize-ii",
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"content": "<p>Given a function <code>fn</code>, return a <strong>memoized</strong> version of that function.</p>\n\n<p>A <strong>memoized </strong>function is a function that will never be called twice with the same inputs. Instead it will return a cached value.</p>\n\n<p><code>fn</code> can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are <code>===</code> to each other.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () => [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\n<strong>Output:</strong> [{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]\n<strong>Explanation:</strong>\nconst inputs = getInputs();\nconst memoized = memoize(fn);\nfor (const arr of inputs) {\n memoized(...arr);\n}\n\nFor the inputs of (2, 2): 2 + 2 = 4, and it required a call to fn().\nFor the inputs of (2, 2): 2 + 2 = 4, but those inputs were seen before so no call to fn() was required.\nFor the inputs of (1, 2): 1 + 2 = 3, and it required another call to fn() for a total of 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () => [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return ({...a, ...b}); }\n<strong>Output:</strong> [{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]\n<strong>Explanation:</strong>\nMerging two empty objects will always result in an empty object. It may seem like there should only be 1 call to fn() because of cache-hits, however none of those objects are === to each other.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\n<strong>Output:</strong> [{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]\n<strong>Explanation:</strong>\nMerging two empty objects will always result in an empty object. The 2nd and 3rd third function calls result in a cache-hit. This is because every object passed in is identical.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= inputs.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= inputs.flat().length <= 10<sup>5</sup></code></li>\n\t<li><code>inputs[i][j] != NaN</code></li>\n</ul>\n",
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"translatedTitle": "记忆函数 II",
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"translatedContent": "<p>请你编写一个函数，它接收一个函数参数 <code>fn</code>，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><code>fn</code> 可以是任何函数，对于它接受什么类型的值没有限制。如果输入为 <code>===</code>，则认为输入相同。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong> \ngetInputs = () => [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\n<b>输出：</b>[{\"val\":4,\"calls\":1},{\"val\":4,\"calls\":1},{\"val\":3,\"calls\":2}]\n<strong>解释：</strong>\nconst inputs = getInputs();\nconst memoized = memoize(fn);\nfor (const arr of inputs) {\n memoized(...arr);\n}\n\n对于参数为 (2, 2) 的输入: 2 + 2 = 4，需要调用 fn() 。\n对于参数为 (2, 2) 的输入: 2 + 2 = 4，这些输入此前调用过，因此不需要调用 fn() 。\n对于参数为 (1, 2) 的输入: 1 + 2 = 3，需要再次调用 fn()，总调用数为 2 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>\ngetInputs = () => [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return a + b; }\n<b>输出：</b>[{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":2},{\"val\":{},\"calls\":3}]\n<strong>解释：</strong>\n合并两个空对象总是会得到一个空对象。因为缓存命中，所以只有 1 次对 fn() 的调用，尽管这些对象之间没有一个是相同的（===）。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong> \ngetInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\n<b>输出：</b>[{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":1}]\n<strong>解释：</strong>\n合并两个空对象总是会得到一个空对象。第 2 和第 3 个函数调用导致缓存命中。这是因为传入的每个对象都是相同的。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= inputs.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= inputs.flat().length <= 10<sup>5</sup></code></li>\n\t<li><code>inputs[i][j] != NaN</code></li>\n</ul>\n",
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"code": "/**\n * @param {Function} fn\n */\nfunction memoize(fn) {\n return function() {\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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"Just because JSON.stringify(obj1) === JSON.stringify(obj2), doesn't necessarily mean obj1 === obj2.",
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"You could iterate over all previously passed inputs to check if there has been a match. However, that will be very slow.",
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"Javascript Maps are a could way to associate arbitrary data.",
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"Make a tree structure of Maps. The depth of the tree should match the number of input parameters."
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