Given an integer array arr and a mapping function fn, return a new array with a transformation applied to each element.
The returned array should be created such that returnedArray[i] = fn(arr[i], i).
Please solve it without the built-in Array.map method.
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Example 1:
\n\n\nInput: arr = [1,2,3], fn = function plusone(n) { return n + 1; }\nOutput: [2,3,4]\nExplanation:\nconst newArray = map(arr, plusone); // [2,3,4]\nThe function increases each value in the array by one. \n\n\nExample 2:
\n\n\nInput: arr = [1,2,3], fn = function plusI(n, i) { return n + i; }\nOutput: [1,3,5]\nExplanation: The function increases each value by the index it resides in.\n\n\nExample 3:
\n\n\nInput: arr = [10,20,30], fn = function constant() { return 42; }\nOutput: [42,42,42]\nExplanation: The function always returns 42.\n\n\n\n
Constraints:
\n\n0 <= arr.length <= 1000-109 <= arr[i] <= 109fn returns a number编写一个函数，这个函数接收一个整数数组 arr 和一个映射函数 fn ，通过该映射函数返回一个新的数组。
返回数组的创建语句应为 returnedArray[i] = fn(arr[i], i) 。
请你在不使用内置方法 Array.map 的前提下解决这个问题。
\n\n
示例 1:
\n\n\n输入：arr = [1,2,3], fn = function plusone(n) { return n + 1; }\n输出：[2,3,4]\n解释： \nconst newArray = map(arr, plusone); // [2,3,4]\n此映射函数返回值是将数组中每个元素的值加 1。\n\n\n示例 2:
\n\n\n输入：arr = [1,2,3], fn = function plusI(n, i) { return n + i; }\n输出：[1,3,5]\n解释：此映射函数返回值根据输入数组索引增加每个值。\n\n\n示例 3:
\n\n\n输入：arr = [10,20,30], fn = function constant() { return 42; }\n输出：[42,42,42]\n解释：此映射函数返回值恒为 42。\n\n\n\n\n
提示：
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