Leetcode简单 乞丐问题 English beggars(JS)

刷题系列

Born a misinterpretation of this kata, your task here is pretty simple: given an array of values and an amount of beggars, you are supposed to return an array with the sum of what each beggar brings home, assuming they all take regular turns, from the first to the last.

For example: [1,2,3,4,5] for 2 beggars will return a result of [9,6], as the first one takes [1,3,5], the second collects [2,4].

The same array with 3 beggars would have in turn have produced a better out come for the second beggar: [5,7,3], as they will respectively take [1,4], [2,5] and [3].

Also note that not all beggars have to take the same amount of “offers”, meaning that the length of the array is not necessarily a multiple of n; length can be even shorter, in which case the last beggars will of course take nothing (0).

Note: in case you don’t get why this kata is about English beggars, then you are not familiar on how religiously queues are taken in the kingdom 😉

Note 2: do not modify the input array.

解题过程：
很简单的一题，当然要想出时间复杂度优秀的解法对于我这样的菜鸟来说还是需要花点时间去思考的，这里使用了js的Array(n)来创建一个长度为n，默认元素为empty的数组，然后用Array.prototype.fill()来把所有元素初始化为0；接下来是关键一步：先定义一个reducer，然后使用reduce() 方法对数组中的每个元素执行reducer函数，将其结果汇总为单个返回值。
reducer 函数接收4个参数:

1. Accumulator (acc) (累计器)
2. Current Value (cur) (当前值)
3. Current Index(idx) (当前索引)
4. Source Array (src) (源数组)

function beggars(values, n){
  //your code here
  var sums = Array(n).fill(0);
  const reducer = (array, amount, index) => {
        array[index % array.length] += amount;
        return array;
    };
  return values.reduce(reducer,sums);
}

题目链接: English beggars.