freeCodeCamp/curriculum/challenges/chinese/22-rosetta-code/rosetta-code-challenges/farey-sequence.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
59c3ec9f15068017c96eb8a3	法里序列	1	302266	farey-sequence

--description--

The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.

Farey 序列 有时被错误地称为 Farey 系列。

每个 Farey 序列：

• starts with the value 0, denoted by the fraction $ \frac{0}{1} $
• ends with the value 1, denoted by the fraction $ \frac{1}{1}$.

Farey 序列的 1 阶到 5 阶是：

• ${\bf\it{F}}_1 = \frac{0}{1}, \frac{1}{1}$
• ${\bf\it{F}}_2 = \frac{0}{1}, \frac{1}{2}, \frac{1}{1}$
• ${\bf\it{F}}_3 = \frac{0}{1}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{1}{1}$
• ${\bf\it{F}}_4 = \frac{0}{1}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{1}{1}$
• ${\bf\it{F}}_5 = \frac{0}{1}, \frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{1}{1}$

--instructions--

编写一个返回 n 阶 Farey 序列的函数。 该函数应该有一个参数，即 n。 It should return the sequence as an array.

--hints--

farey 应该是一个函数。

assert(typeof farey === 'function');

farey(3) 应该返回一个数组

assert(Array.isArray(farey(3)));

farey(3) should return ['0/1','1/3','1/2','2/3','1/1']

assert.deepEqual(farey(3),['0/1', '1/3', '1/2', '2/3', '1/1']);

farey(4) should return ['0/1','1/4','1/3','1/2','2/3','3/4','1/1']

assert.deepEqual(farey(4), ['0/1', '1/4', '1/3', '1/2', '2/3', '3/4', '1/1']);

farey(5) should return ['0/1','1/5','1/4','1/3','2/5','1/2','3/5','2/3','3/4','4/5','1/1']

assert.deepEqual(farey(5), [
  '0/1',
  '1/5',
  '1/4',
  '1/3',
  '2/5',
  '1/2',
  '3/5',
  '2/3',
  '3/4',
  '4/5',
  '1/1'
]);

--seed--

--seed-contents--

function farey(n) {

}

--solutions--

function farey(n) {
  const sequence = [{ string: "0/1", float: 0.0 }];
  for (let i = 1; i < n; i++) {
    for (let j = n; j >= i; j--) {
      if (i === 1 || j % i > 0) {
        sequence.push({ string: `${i}/${j}`, float: i / j });
      }
    }
  }
  return sequence
    .sort((a, b) => a.float - b.float)
    .map(e => e.string)
}