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

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            "title": "Memoize II",
            "titleSlug": "memoize-ii",
            "content": "<p>Given a function <code>fn</code>,&nbsp;return&nbsp;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 return&nbsp;a cached value.</p>\n\n<p><code>fn</code>&nbsp;can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are&nbsp;<code>===</code> to each other.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () =&gt; [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\n<strong>Output:</strong> [{&quot;val&quot;:4,&quot;calls&quot;:1},{&quot;val&quot;:4,&quot;calls&quot;:1},{&quot;val&quot;:3,&quot;calls&quot;: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 = () =&gt; [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return ({...a, ...b}); }\n<strong>Output:</strong> [{&quot;val&quot;:{},&quot;calls&quot;:1},{&quot;val&quot;:{},&quot;calls&quot;:2},{&quot;val&quot;:{},&quot;calls&quot;: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&nbsp;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 = () =&gt; { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\n<strong>Output:</strong> [{&quot;val&quot;:{},&quot;calls&quot;:1},{&quot;val&quot;:{},&quot;calls&quot;:1},{&quot;val&quot;:{},&quot;calls&quot;: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>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= inputs.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= inputs.flat().length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>inputs[i][j] != NaN</code></li>\n</ul>\n",
            "translatedTitle": "记忆函数 II",
            "translatedContent": "<p>现给定一个函数 <code>fn</code> ，返回该函数的一个 <strong>记忆化</strong> 版本。</p>\n\n<p>一个 <strong>记忆化</strong> 的函数是一个函数，它不会被相同的输入调用两次。而是会返回一个缓存的值。</p>\n\n<p>函数 <code>fn</code> 可以是任何函数，对它所接受的值类型没有任何限制。如果两个输入值在 JavaScript 中使用 <code>===</code>&nbsp;运算符比较时相等，则它们被视为相同。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong> \ngetInputs = () =&gt; [[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，这些输入之前已经出现过，因此不需要再次调用 fn()。\n对于参数为 (1, 2) 的输入: 1 + 2 = 3，需要再次调用 fn()，总共调用了 2 次。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>\ngetInputs = () =&gt; [[{},{}],[{},{}],[{},{}]] \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将两个空对象合并总是会得到一个空对象。尽管看起来应该缓存命中并只调用一次 <code>fn()</code>，但是这些空对象彼此之间都不是 <code>===</code> 相等的。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong> \ngetInputs = () =&gt; { 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将两个空对象合并总是会得到一个空对象。因为传入的每个对象都是相同的，所以第二个和第三个函数调用都会命中缓存。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= inputs.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= inputs.flat().length &lt;= 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 * @return {Function}\n */\nfunction memoize(fn) {\n    \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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                    "code": "type Fn = (...params: any) => any\n\nfunction memoize(fn: Fn): Fn {\n    \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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            "hints": [
                "Just because JSON.stringify(obj1) === JSON.stringify(obj2), doesn't necessarily mean obj1 === obj2.",
                "You could iterate over all previously passed inputs to check if there has been a match. However, that will be very slow.",
                "Javascript Maps are a could way to associate arbitrary data.",
                "Make a tree structure of Maps. The depth of the tree should match the number of input parameters."
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