Oliver Eyton-Williams ee1e8abd87

feat(curriculum): restore seed + solution to Chinese (#40683 )

* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>

2021-01-12 19:31:00 -07:00

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id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
afd15382cdfb22c9efe8b7de	DNA 配对	5	16009	dna-pairing

--description--

给出的 DNA 链上缺少配对元素。请基于每个字符，获取与其配对的元素，并将结果作为二维数组返回。

DNA 的碱基对有两种形式：一种是 A 与 T，一种是 C 与 G。请为参数中给出的每个字符配对相应的碱基。

注意，参数中给出的字符应作为每个子数组中的第一个元素返回。

例如，传入 GCG 时，应返回 "G", "C"], ["C", "G"], ["G", "C"。

参数中的字符及与其配对的碱基应存在于一个数组中，代表碱基对。再将每个配对完成的碱基对数组按顺序放到一个数组中，作为最终的返回结果。

--hints--

pairElement("ATCGA") 应返回 [["A","T"],["T","A"],["C","G"],["G","C"],["A","T"]]。

assert.deepEqual(pairElement('ATCGA'), [
  ['A', 'T'],
  ['T', 'A'],
  ['C', 'G'],
  ['G', 'C'],
  ['A', 'T']
]);

pairElement("TTGAG") 应返回 [["T","A"],["T","A"],["G","C"],["A","T"],["G","C"]]。

assert.deepEqual(pairElement('TTGAG'), [
  ['T', 'A'],
  ['T', 'A'],
  ['G', 'C'],
  ['A', 'T'],
  ['G', 'C']
]);

pairElement("CTCTA") 应返回 [["C","G"],["T","A"],["C","G"],["T","A"],["A","T"]]。

assert.deepEqual(pairElement('CTCTA'), [
  ['C', 'G'],
  ['T', 'A'],
  ['C', 'G'],
  ['T', 'A'],
  ['A', 'T']
]);

--seed--

--seed-contents--

function pairElement(str) {
  return str;
}

pairElement("GCG");

--solutions--

var lookup = Object.create(null);
lookup.A = 'T';
lookup.T = 'A';
lookup.C = 'G';
lookup.G = 'C';

function pairElement(str) {
 return str.split('').map(function(p) {return [p, lookup[p]];});
}

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