给定一个函数 fn，返回该函数的记忆版本。
memoized 函数是永远不会使用相同输入调用两次的函数。相反，它将返回一个缓存值。
您可以假设有 3 种可能的输入函数：sum、fib 和阶乘。
sum 接受两个整数 a 和 b 并返回 a b。假设如果已经为参数 (b, a) 缓存了一个值，其中 a != b，则该值不能用于参数 (a, b)。例如，如果参数是 (3, 2) 和 (2, 3)，则应进行两次单独的调用。
fib 接受单个整数 n，如果 n 阶乘接受单个整数 n，如果 n
示例1：
输入：
fnName = "sum" actions = ["call","call","getCallCount","call","getCallCount"] values = [[2,2],[2,2],[],[1,2],[]]
输出： [4,4,1,3,2]
说明：
const sum = (a, b) => a + b; const memoizedSum = memoize(sum); memoizedSum(2, 2); // "call" - returns 4. sum() was called as (2, 2) was not seen before. memoizedSum(2, 2); // "call" - returns 4. However sum() was not called because the same inputs were seen before. // "getCallCount" - total call count: 1 memoizedSum(1, 2); // "call" - returns 3. sum() was called as (1, 2) was not seen before. // "getCallCount" - total call count: 2
示例2：
输入：
fnName = "factorial" actions = ["call","call","call","getCallCount","call","getCallCount"] values = [[2],[3],[2],[],[3],[]]
输出： [2,6,2,2,6,2]
说明：
const factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1)); const memoFactorial = memoize(factorial); memoFactorial(2); // "call" - returns 2. memoFactorial(3); // "call" - returns 6. memoFactorial(2); // "call" - returns 2. However factorial was not called because 2 was seen before. // "getCallCount" - total call count: 2 memoFactorial(3); // "call" - returns 6. However factorial was not called because 3 was seen before. // "getCallCount" - total call count: 2
示例 3：
输入：
fnName = "fib" actions = ["call","getCallCount"] values = [[5],[]]
输出： [8,1]
说明：
fib(5) = 8 // "call" // "getCallCount" - total call count: 1
约束：
> 0 <= a, b <= 105 > 1 <= n <= 10 > 0 <= actions.length <= 105 > actions.length === values.length
actions[i] 是“call”和“getCallCount”之一
fnName 是“sum”、“factorial”和“fib”之一
当您继续开发和优化 JavaScript 应用程序时，请记住记忆的力量。通过确定正确的函数来记忆并实施适当的缓存策略，您可以显着提高性能，并为您的客户创造更加无缝和响应更快的用户体验。
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