使用递归替换循环

Question

我真的需要反馈我的解释，尤其是关于2(和3(的解释。我只是想确认我是否理解正确，因为我是一个新手
这是一个免费的CodeCamp挑战，对我来说真的很有挑战性，因为我以前没有JS的经验。如下所示
编写一个递归函数sum(arr，n(，返回数组arr的前n个元素的和。

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
*/

Answer 1

从逻辑视图来看，它看起来不错。如果你想玩的话，你也可以试试三元运算符。

return n <= 0 ? 0 : sum(arr, n - 1) + arr[n - 1];

第一块是if问题。如果它的true为，则转到第二个块(从whith？开始(；如果它的false为，则转至第三个块(以：开始(。

Answer 2

我认为我们不能缩短它：

const sum = (arr,n) => --n<0 ? 0 : sum(arr,n) +arr[n]
console.log ( sum([1], 0) )
console.log ( sum([2, 3, 4], 1) )
console.log ( sum([2, 3, 4, 5], 3) )

.as-console-wrapper { max-height: 100% !important; top: 0; }