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\nYou can assume there are 3 possible input functions: sum, fib, and factorial.
sum accepts two integers a and b and returns a + b.fib accepts a single integer n and returns 1 if n <= 1 or fib(n - 1) + fib(n - 2) otherwise.factorial accepts a single integer n and returns 1 if n <= 1 or factorial(n - 1) * n otherwise.\n
Example 1:
\n\n\nInput:\nfnName = "sum"\nactions = ["call","call","getCallCount","call","getCallCount"]\nvalues = [[2,2],[2,2],[],[1,2],[]]\nOutput: [4,4,1,3,2]\nExplanation:\nconst sum = (a, b) => a + b;\nconst memoizedSum = memoize(sum);\nmemoizedSum(2, 2); // "call" - returns 4. sum() was called as (2, 2) was not seen before.\nmemoizedSum(2, 2); // "call" - returns 4. However sum() was not called because the same inputs were seen before.\n// "getCallCount" - total call count: 1\nmemoizedSum(1, 2); // "call" - returns 3. sum() was called as (1, 2) was not seen before.\n// "getCallCount" - total call count: 2\n\n\n
Example 2:
\n\n\nInput:\nfnName = "factorial"\nactions = ["call","call","call","getCallCount","call","getCallCount"]\nvalues = [[2],[3],[2],[],[3],[]]\nOutput: [2,6,2,2,6,2]\nExplanation:\nconst factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1));\nconst memoFactorial = memoize(factorial);\nmemoFactorial(2); // "call" - returns 2.\nmemoFactorial(3); // "call" - returns 6.\nmemoFactorial(2); // "call" - returns 2. However factorial was not called because 2 was seen before.\n// "getCallCount" - total call count: 2\nmemoFactorial(3); // "call" - returns 6. However factorial was not called because 3 was seen before.\n// "getCallCount" - total call count: 2\n\n\n
Example 3:
\n\n\nInput:\nfnName = "fib"\nactions = ["call","getCallCount"]\nvalues = [[5],[]]\nOutput: [8,1]\nExplanation:\nfib(5) = 8 // "call"\n// "getCallCount" - total call count: 1\n\n\n
\n
Constraints:
\n\n0 <= a, b <= 1051 <= n <= 100 <= actions.length <= 105actions.length === values.lengthactions[i] is one of "call" and "getCallCount"fnName is one of "sum", "factorial" and "fib"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 记忆化 后的结果。
\n\n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\n你可以假设有 3 个可能的输入函数：sum 、fib 和 factorial 。
sum 接收两个整型参数 a 和 b ，并返回 a + b 。fib 接收一个整型参数 n ，如果 n <= 1 则返回 1，否则返回 fib (n - 1) + fib (n - 2)。factorial 接收一个整型参数 n ，如果 n <= 1 则返回 1 ，否则返回 factorial(n - 1) * n 。\n\n
示例 1：
\n\n\n输入：\nfnName = \"sum\"\nactions = [\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\nvalues = [[2,2],[2,2],[],[1,2],[]]\n输出：[4,4,1,3,2]\n解释：\nconst sum = (a, b) => a + b;\nconst memoizedSum = memoize(sum);\nmemoizedSum (2, 2);// \"call\" - 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);// \"call\" - 返回 4。没有调用 sum()，因为前面有相同的输入。\n// \"getCallCount\" - 总调用数： 1\nmemoizedSum(1、2);// \"call\" - 返回 3。sum() 被调用，因为之前没有使用参数 (1, 2) 调用过。\n// \"getCallCount\" - 总调用数： 2\n\n\n
示例 2：
\n\n\n输入：\nfnName = \"factorial\"\nactions = [\"call\",\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\nvalues = [[2],[3],[2],[],[3],[]]\n输出：[2,6,2,2,6,2]\n解释：\nconst factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1));\nconst memoFactorial = memoize(factorial);\nmemoFactorial(2); // \"call\" - 返回 2。\nmemoFactorial(3); // \"call\" - 返回 6。\nmemoFactorial(2); // \"call\" - 返回 2。 没有调用 factorial()，因为前面有相同的输入。\n// \"getCallCount\" - 总调用数：2\nmemoFactorial(3); // \"call\" - 返回 6。 没有调用 factorial()，因为前面有相同的输入。\n// \"getCallCount\" - 总调用数：2\n\n\n
示例 3：
\n\n\n输入：\nfnName = \"fib\"\nactions = [\"call\",\"getCallCount\"]\nvalues = [[5],[]]\n输出：[8,1]\n解释：\nfib(5) = 8 // \"call\"\n// \"getCallCount\" - 总调用数：1\n\n\n\n
\n\n
提示：
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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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