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125 lines
2.8 KiB
Markdown
125 lines
2.8 KiB
Markdown
---
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id: 594faaab4e2a8626833e9c3d
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title: 使用转义标记字符串
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challengeType: 1
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forumTopicId: 302338
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dashedName: tokenize-a-string-with-escaping
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---
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# --description--
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Write a function or program that can split a string at each non-escaped occurrence of a separator character.
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它应该接受三个输入参数：
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<ul>
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<li>The <strong>string</strong></li>
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<li>The <strong>separator character</strong></li>
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<li>The <strong>escape character</strong></li>
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</ul>
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它应该输出一个字符串列表。
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拆分规则：
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<ul>
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<li>The fields that were separated by the separators, become the elements of the output list.</li>
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<li>即使在开始和结束时，也应保留空字段。</li>
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</ul>
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转义规则：
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<ul>
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<li>"Escaped" means preceded by an occurrence of the escape character that is not already escaped itself.</li>
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<li>当转义字符位于没有特殊含义的字符之前时，它仍然算作转义字符（但不会做任何特殊的事情）。</li>
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<li>用于转义某些内容的每次出现的转义字符不应成为输出的一部分。</li>
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</ul>
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证明您的函数满足以下测试用例：
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给定字符串
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<pre>one^|uno||three^^^^|four^^^|^cuatro|</pre>
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并使用 `|` 作为分隔符和 `^` 作为转义字符，你的函数应输出以下数组：
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<pre> ['one|uno', '', 'three^^', 'four^|cuatro', '']
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</pre>
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# --hints--
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`tokenize` 应该是一个函数。
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```js
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assert(typeof tokenize === 'function');
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```
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`tokenize` 应该返回一个数组。
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```js
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assert(typeof tokenize('a', 'b', 'c') === 'object');
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```
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`tokenize('one^|uno||three^^^^|four^^^|^cuatro|', '|', '^')` 应该返回 `['one|uno', '', 'three^^', 'four^|cuatro', '']`
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```js
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assert.deepEqual(tokenize(testStr1, '|', '^'), res1);
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```
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`tokenize('a@&bcd&ef&&@@hi', '&', '@')` 应该返回 `['a&bcd', 'ef', '', '@hi']`
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```js
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assert.deepEqual(tokenize(testStr2, '&', '@'), res2);
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```
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# --seed--
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## --after-user-code--
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```js
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const testStr1 = 'one^|uno||three^^^^|four^^^|^cuatro|';
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const res1 = ['one|uno', '', 'three^^', 'four^|cuatro', ''];
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// TODO add more tests
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const testStr2 = 'a@&bcd&ef&&@@hi';
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const res2 = ['a&bcd', 'ef', '', '@hi'];
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```
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## --seed-contents--
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```js
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function tokenize(str, sep, esc) {
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return true;
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}
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```
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# --solutions--
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```js
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// tokenize :: String -> Character -> Character -> [String]
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function tokenize(str, charDelim, charEsc) {
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const dctParse = str.split('')
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.reduce((a, x) => {
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const blnEsc = a.esc;
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const blnBreak = !blnEsc && x === charDelim;
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const blnEscChar = !blnEsc && x === charEsc;
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return {
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esc: blnEscChar,
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token: blnBreak ? '' : (
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a.token + (blnEscChar ? '' : x)
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),
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list: a.list.concat(blnBreak ? a.token : [])
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};
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}, {
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esc: false,
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token: '',
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list: []
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});
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return dctParse.list.concat(
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dctParse.token
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);
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}
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```
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