leetcode 2. 给定一个整数n，返回一个计数器函数。此计数器函数最初返回n，然后每次调用时（n、n+1、n+2等）都会比前一个值多返回1。

给定一个整数n，返回一个计数器函数。此计数器函数最初返回n，然后每次调用时（n、n+1、n+2等）都会比前一个值多返回1。

Example 1:

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.
Example 2:

Input: 
n = -2
["call","call","call","call","call"]
Output: [-2,-1,0,1,2]
Explanation: counter() initially returns -2. Then increases after each sebsequent call.
 

Constraints:

-1000 <= n <= 1000
0 <= calls.length <= 1000
calls[i] === "call"

解决

function createCounter(n: number): () => number {
    return function() {
        return n++;        
    }
}


/** 
 * const counter = createCounter(10)
 * counter() // 10
 * counter() // 11
 * counter() // 12
 */

知识点： 利用闭包（closure）来记录函数被执行的次数
1.先定义一个普通函数：（列如这里的求和）

2.再在闭包中创建累加器并包裹这个函数， 并给原函数重新赋值

3. 测试

   

   4.结果