# 2665.Counter II

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

## 題目
Write a function createCounter. It should accept an initial integer init. It should return an object with three functions.

The three functions are:

- increment() increases the current value by 1 and then returns it.
- decrement() reduces the current value by 1 and then returns it.
- reset() sets the current value to init and then returns it.

## Example

Input: init = 5, calls = ["increment","reset","decrement"]
Output: [6,5,4]
Explanation:
const counter = createCounter(5);
counter.increment(); // 6
counter.reset(); // 5
counter.decrement(); // 4

Input: init = 0, calls = ["increment","increment","decrement","reset","reset"]
Output: [1,2,1,0,0]
Explanation:
const counter = createCounter(0);
counter.increment(); // 1
counter.increment(); // 2
counter.decrement(); // 1
counter.reset(); // 0
counter.reset(); // 0

### Constraints:
- -1000 <= init <= 1000
- total calls not to exceed 1000

---
## 解題邏輯
這題在 createCounter 這個函式裡有一個 init 的參數，以及回傳三個函式分別為`incretment`,`reset`,`decrement`，並且能夠以

const counter = createCounter (5)
counter.increpment()

的方式呼叫三個函式
### 先來第一版直覺想法

var createCounter = function(init) {
  let allFn = {}
  const increment = ()=>{
 return init +1
  }
    const reset = ()=>{
 return init
  }
    const decrement = ()=>{
 return init -1
  }
   allFn.increment = increment 
   allFn. reset =  reset
   allFn.decrement= decrement
    return allFn
  }
 

第一版跑第一個側向是通過的

const counter = createCounter(5)
counter.increment();//6
counter.reset();//5
counter.decrement();//4

但跑第二個就壞了ＱＱ

const counter = createCounter(0)
counter.increment(); // 1
counter.increment(); // 1
counter.decrement(); // -1
counter.reset(); // 0
counter.reset(); // 0

可以看到在第二個側向中每一次的加減法都是拿`init`這個數去加減，彼此之間沒有任何關係，但並不符合題目，題目要的是下一個函式的`init`都是接續上一個函式的結果而來的，思考後發現只要將加、減之間所引用的參數加上關聯，`reset`這個函式就是回到最初的`init`沒問題

var createCounter = function(init) {
  let num = init
  let allFn = {}
  const increment = ()=>{
 return num = num+1
  }
    const reset = ()=>{
 return num = init
  }
    const decrement = ()=>{
 return num = num-1
  }
   allFn.increment = increment 
   allFn. reset =  reset
   allFn.decrement= decrement
    return allFn
  }
 

### 將 init 賦值新變數
因此讓`increment`,`decrement`的數字來源都是同一個新變數`num`，這樣就能確保都是用同一個數字作加減，也能將結果延續了，參考解說後發現 object 的寫法可以再優化

var createCounter = function(init) {
  let num = init
    return {
        increment : function(){
            return num =num+1
                },
        reset : function() {
            return num = init
                },
        decrement : function (){
            return num=num-1
                }
        }
  }
 

### 終極優化版

var createCounter = function(init) {
  let num = init
    return {
        increment : ()=>num =num+1,
        reset :()=>num =init,
        decrement : ()=> num=num-1
    }
  }
 

{%hackmd @themes/dracula %}