Given a function fn, return a memoized version of that function.
A memoized function is a function that will never be called twice with the same inputs. Instead it will return a cached value.
\n\nfn can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are === to each other.
\n
Example 1:
\n\n\nInput: \ngetInputs = () => [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\nOutput: [{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]\nExplanation:\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\n\nExample 2:
\n\n\nInput: \ngetInputs = () => [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return ({...a, ...b}); }\nOutput: [{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]\nExplanation:\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\n\nExample 3:
\n\n\nInput: \ngetInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\nOutput: [{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]\nExplanation:\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\n\n\n
Constraints:
\n\n1 <= inputs.length <= 1050 <= inputs.flat().length <= 105inputs[i][j] != NaN请你编写一个函数，它接收一个函数参数 fn，并返回该函数的 记忆化 后的结果。
记忆函数 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。
\n\nfn 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 ===，则认为输入相同。
\n\n
示例 1：
\n\n\n输入： \ngetInputs = () => [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\n输出：[{\"val\":4,\"calls\":1},{\"val\":4,\"calls\":1},{\"val\":3,\"calls\":2}]\n解释：\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\n\n示例 2：
\n\n\n输入：\ngetInputs = () => [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return a + b; }\n输出：[{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":2},{\"val\":{},\"calls\":3}]\n解释：\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\n\n示例 3：
\n\n\n输入： \ngetInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\n输出：[{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":1}]\n解释：\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\n\n\n\n
提示：
\n\n1 <= inputs.length <= 1050 <= inputs.flat().length <= 105inputs[i][j] != NaN\\u7248\\u672c\\uff1a \\u60a8\\u7684\\u4ee3\\u7801\\u5728\\u6267\\u884c\\u65f6\\u5c06\\u5e26\\u4e0a lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 datastructures-js\\/priority-queue@5.3.0<\\/a> \\u548c datastructures-js\\/queue@4.2.1<\\/a>\\u3002<\\/p>\"],\"typescript\":[\"TypeScript\",\" TypeScript 4.5.4<\\/p>\\r\\n\\r\\n Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2020<\\/p>\\r\\n\\r\\n lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 datastructures-js\\/priority-queue@5.3.0<\\/a> \\u548c datastructures-js\\/queue@4.2.1<\\/a>\\u3002<\\/p>\"]}",
"book": null,
"isSubscribed": false,
"isDailyQuestion": false,
"dailyRecordStatus": null,
"editorType": "CKEDITOR",
"ugcQuestionId": null,
"style": "LEETCODE",
"exampleTestcases": "() => [[2,2],[2,2],[1,2]]\nfunction (a, b) { return a + b; }\n() => [[{},{}],[{},{}],[{},{}]]\nfunction (a, b) { return ({...a, ...b}); }\n() => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfunction (a, b) { return ({...a, ...b}); }",
"__typename": "QuestionNode"
}
}
}Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\n