###### tags:`Function` | `leetCode`
<font color="#01AE9A" background="E1F3F0">`easy`</font>

### 題目
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.

You 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.

### Example

Input
"sum"
["call","call","getCallCount","call","getCallCount"]
[[2,2],[2,2],[],[1,2],[]]
Output
[4,4,1,3,2]

Explanation
const sum = (a, b) => a + b;
const memoizedSum = memoize(sum);
memoizedSum(2, 2); // Returns 4. sum() was called as (2, 2) was not seen before.
memoizedSum(2, 2); // Returns 4. However sum() was not called because the same inputs were seen before.
// Total call count: 1
memoizedSum(1, 2); // Returns 3. sum() was called as (1, 2) was not seen before.
// Total call count: 2

---
### 解題邏輯
這題光是題目就有點難理解了～基本上就是寫一個會記憶的 memoize fn,如果傳給他[2,2],他會return 2+2=4,如果再傳入一樣的[2,2],則要回傳剛剛存起來的結果而不是重新計算一次

 let callCount = 0;
 const memoizedFn = memoize(function (a, b) {
	 callCount += 1;
   return a + b;
 })
memoizedFn(2, 3) // 5
memoizedFn(2, 3) // 5
console.log(callCount) // 1 

function memoize(fn) {
    const cache = {}
    // 設一個 cache 的空物件
    return function(...args) {
      const key =  JSON.stringify(args)
      // 把傳進來的參數轉為字串（因為如果比對[2,2]===[2,2],因為兩個 arr 所存的記憶體不同所以會回傳 false）
      if(key in cache){
          // 如果 cache 中有一樣的 key 了就直接回傳 cache[key]
        return cache[key]
      }
      const output = fn(...args)
      // 把 output 指定給 cache 裡的 key
      cache[key] =  output
      return output
    }
}