* 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>
2.7 KiB
2.7 KiB
id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
594faaab4e2a8626833e9c3d | 使用转义标记字符串 | 5 | tokenize-a-string-with-escaping |
--description--
编写一个函数或程序,可以在分隔符的每个非转义事件中拆分字符串。
它应该接受三个输入参数:
字符串 分隔符字符 转义字符它应该输出一个字符串列表。
拆分规则:
由分隔符分隔的字段将成为输出列表的元素。应保留空字段,即使在开始和结束时也是如此。转义规则:
“Escaped”意味着出现一个尚未自行转义的转义字符。当转义字符位于没有特殊含义的字符之前时,它仍然被视为转义符(但不会做任何特殊操作)。用于转义某些内容的每次出现的转义字符都不应成为输出的一部分。证明您的函数满足以下测试用例:给定字符串
一个^ | UNO || 3 ^^^^ |四^^^ | ^夸|和使用
|作为分隔符和
^作为转义字符,您的函数应输出以下数组:
['one | uno',“,'three ^^','four ^ | quatro',”]
--hints--
tokenize
是一个函数。
assert(typeof tokenize === 'function');
tokenize
应该返回一个数组。
assert(typeof tokenize('a', 'b', 'c') === 'object');
tokenize("one^|uno||three^^^^|four^^^|^cuatro|", "|", "^")
应返回[“one | uno”,“”,“three ^^” ,“四个^ | cuatro”,“”]“)
assert.deepEqual(tokenize(testStr1, '|', '^'), res1);
tokenize("a@&bcd&ef&&@@hi", "&", "@")
应返回["a&bcd", "ef", "", "@hi"]
assert.deepEqual(tokenize(testStr2, '&', '@'), res2);
--seed--
--after-user-code--
const testStr1 = 'one^|uno||three^^^^|four^^^|^cuatro|';
const res1 = ['one|uno', '', 'three^^', 'four^|cuatro', ''];
// TODO add more tests
const testStr2 = 'a@&bcd&ef&&@@hi';
const res2 = ['a&bcd', 'ef', '', '@hi'];
--seed-contents--
function tokenize(str, sep, esc) {
return true;
}
--solutions--
// tokenize :: String -> Character -> Character -> [String]
function tokenize(str, charDelim, charEsc) {
const dctParse = str.split('')
.reduce((a, x) => {
const blnEsc = a.esc;
const blnBreak = !blnEsc && x === charDelim;
const blnEscChar = !blnEsc && x === charEsc;
return {
esc: blnEscChar,
token: blnBreak ? '' : (
a.token + (blnEscChar ? '' : x)
),
list: a.list.concat(blnBreak ? a.token : [])
};
}, {
esc: false,
token: '',
list: []
});
return dctParse.list.concat(
dctParse.token
);
}