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