使用递归替换循环

Question

我真的需要反馈我的解释，尤其是 2) 和 3)。我只是想确认一下我是否理解正确，因为我是新手。

这是一个 freeCodeCamp 挑战，对我来说确实很有挑战性，因为我以前没有 JS 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 个元素的总和。

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

从逻辑角度来看，它看起来不错。如果你想玩一下，你也可以尝试三元运算符。return&nbsp;n&nbsp;<=&nbsp;0&nbsp;?&nbsp;0&nbsp;:&nbsp;sum(arr,&nbsp;n&nbsp;-&nbsp;1)&nbsp;+&nbsp;arr[n&nbsp;-&nbsp;1];第一个块是 if 问题。如果为真，则转到第二个块（以 : 开头），如果为假，则转到第三个块（以 : 开头）。

Answer 2

我认为我们无法缩短它：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; }