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：（比如这里的求和）
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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
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测试
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4.结果
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