# 2620.Counter

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

### 題目
Given an integer n, return a counter function. This counter function initially returns n and then returns 1 more than the previous value every subsequent time it is called (n, n + 1, n + 2, etc).
### Example

Input: 
n = 10 
["call","call","call"]
Output: [10,11,12]
Explanation: 
counter() = 10 // The first time counter() is called, it returns n.
counter() = 11 // Returns 1 more than the previous time.
counter() = 12 // Returns 1 more than the previous time.

### Constraints:
- `-1000 <= n <= 1000`
- `At most 1000 calls to counter() will be made`

---
### 解題邏輯
這題是閉包的題目，關於閉包的概念大概就是==讓 function 內部具私有變數，而不被全域污染==,所以我想把 n+1存給n,再return n

var createCounter = function(n) {
        return function() {
        n = n+1
      return n-1
    };
};

答案雖然有對但是總覺得 return n-1好像怪怪的，所以想了一下，結果有更簡潔的方式

var createCounter = function(n) {
        return function() {
      return n++
    };
};
 const counter = createCounter(10)
counter()//10
counter()//11 
counter()//12

#### 容易搞混的遞增運算子
這邊要注意 n++ 使用了==遞增運算子==，放在 n 後面會先回傳 n 本身，再進行遞增，反之在前綴則會先進行遞增，再將遞增後的結果回傳。

let x = 3;
y = x++;

console.log("x",x)// 4 ->x被遞增1了
console.log("y",y)// 3 y是x沒被遞增的初始值

let x = 3;
y = ++x;

console.log("x",x)// 4 ->x被遞增1後回傳
console.log("y",y)// 4 y是x被遞增後的值

### 參考連結
[PJ大大的閉包筆記-[JS] 深入淺出 JavaScript 閉包（closure）](https://pjchender.dev/javascript/js-closure/)

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