擲出前兩天學到的閉包來解決這一關卡(►__◄)

Day04 - 2665.Counter II EASY

Description❓

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.

宣告createCounter函示，該函示先接受初始整數init作為參數，它應該返回有三個函示的物件。

• increment() 將當前的值+1然後返回
• decrement() 將當前的值-1然後返回
• reset() 將當前的值重置為init後返回

Points

• 30Days of JS 工具書－閉包
• LeetCode JS30-Day02|學習閉包closure
• LeetCode JS30-Day03|設置閉包方法
• 前置遞增運算子

Solution✍️

[ ▶️挑戰這一題 ] [ 本日代碼 ]

1. 這一題由於調用init()時需要return 參數i，所以在外部作用域宣告一個變數init把i初始值存起來。
2. 依照增減的邏輯使用前置遞增運算子，讓返回的值是新的數值。

createCounter=(i)=>{
   let init = i;
   return {
      increment:()=>++i,
      decrement:()=>--i,
      reset:()=>{
         i=init;
         return i
      },
   }
}

Testcase

let counter = createCounter(5);
let case1 = [
   counter.increment(),
   counter.reset(),
   counter.decrement()
]
console.log(case1);// [6, 5, 4]

counter = createCounter(0);
let case2 = [
   counter.increment(),
   counter.increment(),
   counter.decrement(),
   counter.reset(),
   counter.reset(),
];
console.log(case2);// [1, 2, 1, 0, 0]