我真的需要反饋我的解釋，尤其是 2) 和 3)。我只是想確認(rèn)一下我是否理解正確，因?yàn)槲沂切率?。這是一個(gè) freeCodeCamp 挑戰(zhàn)，對(duì)我來(lái)說(shuō)確實(shí)很有挑戰(zhàn)性，因?yàn)槲乙郧皼](méi)有 JS 經(jīng)驗(yàn)。如下所示。編寫(xiě)一個(gè)遞歸函數(shù) sum(arr, n)，返回?cái)?shù)組 arr 的前 n 個(gè)元素的總和。function sum(arr, n) {  if(n <= 0) {    return 0;  } else {    return sum(arr, n - 1) + arr[n - 1];  }}/* 1) sum([1], 0) should equal 0.   2) sum([2, 3, 4], 1) should equal 2.   3) sum([2, 3, 4, 5], 3) should equal 9. *//* My explanations are down below *//*    Explanation 1)     sum([1], 0) should equal 0.    n is less or equal to 0 so line 2 works and returns 0 at line 3.*//*  Explanation 2)     sum([2, 3, 4], 1) should equal 2    n is not less or equal to 0 so it will not return 0 according to line 2. We move to line 5.    *return sum(arr, n - 1) + arr[n - 1];=>  return sum([2, 3, 4], 1 - 1) + arr[1 - 1];=>  return sum([2, 3, 4], 0) + arr[0] => n is less or equal to 0 so it will return zero according to line 2.=>  return 0 + arr[0]=>  Since arr[0] is equals to 2=>  return 0 + 2;=>  2*//*  Explanation 3)    sum([2, 3, 4, 5], 3) should equal 9    n is not less or equal to 0 so it will not return 0 according to line 2. We move to line 5.    *return sum(arr, n - 1) + arr[n - 1];=>  return sum([2, 3, 4, 5], 3 - 1) + arr[3 - 1];=>  return sum([2, 3, 4, 5], 2) + arr[2]; => n is 2, not less or equal to 0 so go back to line 5 + arr[2].=>  return sum(arr, n - 1) + arr[n - 1] + arr[2];=>  return sum([2, 3, 4, 5], 2 - 1) + arr[2 - 1] + arr[2];=>  return sum([2, 3, 4, 5], 1) + arr[1] + arr[2]; => n is 1, not less or equal to 0 so goes back to line 5 arr[1] + arr[2].=>  return sum(arr, n - 1) + arr[n - 1] + arr[1] + arr[2];=>  return sum([2, 3, 4, 5], 1 - 1) + arr[1 - 1] + arr[1] + arr[2];=>  return sum([2, 3, 4, 5], 0) + arr [0] + arr[1] + arr[2];=>  return 0 + arr[0] + arr[1] + arr[2];=>  in our array, arr[0] is 2, arr[1] is 3, arr[2] is 4.=>  0 + 2 + 3 + 4=>  5 + 4=>  returns 9*/