按正态分布排序数字数组(高斯分布)

Question

有一个数字数组,setOfNumbers = [0, 3, 3, 2, 7, 1, -2, 9]我想对这个数组进行排序,使得最后和开始时的数字最小,并且排序集合中心的最大数字就像这样sortedSetNumbers = [0, 2, 3, 9, 7, 3, 1, -2].

const setOfNumbers = [0, 3, 3, 2, 7, 1, -2, 9];
const result = [0, 2, 3, 9, 7, 3, 1, -2];

function sortNormal(a, b) {
  return true; // Please, change this line
}

const sortedSetNumbers = setOfNumbers.sort((a, b) => sortNormal(a, b));



if (sortedSetNumbers === result) {
  console.info('Succeeded Normal Distributed');
} else {
  console.warn('Failed Normal Distribution');
}

console.log(sortedSetNumbers);

我确信可以使用该方法对这些数字进行排序Array.prototype.sort(),但这个排序函数应该如何？

编辑:解决方案不必解决.sort().那只是一个想法.

Answer 1

这可能是最天真的方式,但它不仅仅是左,右,左,右......排序后？

const input    = [0, 3, 3, 2, 7, 1, -2, 9];
const expected = [0, 2, 3, 9, 7, 3, 1, -2];

const sorted   = input.slice().sort();
const output   = [];
let side       = true;

while (sorted.length) {
  output[side ? 'unshift' : 'push'](sorted.pop());
  side = !side;
}

console.log(expected.join());
console.log(output.join());

或者干脆:

const input  = [0, 3, 3, 2, 7, 1, -2, 9];
const output = input.slice().sort().reduceRight((acc, val, i) => {
  return i % 2 === 0 ? [...acc, val] : [val, ...acc];
}, []);

console.log(output.join());