ee1e8abd87
* 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>
79 lines
1.6 KiB
Markdown
79 lines
1.6 KiB
Markdown
---
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id: 596fda99c69f779975a1b67d
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title: 计算子字符串的出现次数
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challengeType: 5
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videoUrl: ''
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dashedName: count-occurrences-of-a-substring
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---
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# --description--
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任务：
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创建函数或显示内置函数，以计算字符串中子字符串的非重叠出现次数。
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该函数应该有两个参数：
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第一个参数是要搜索的字符串，第二个参数是要搜索的子字符串。
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它应该返回一个整数计数。
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匹配应产生最多数量的非重叠匹配。
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通常，这实质上意味着从左到右或从右到左匹配。
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# --hints--
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`countSubstring`是一个函数。
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```js
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assert(typeof countSubstring === 'function');
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```
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`countSubstring("the three truths", "th")`应该返回`3` 。
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```js
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assert.equal(countSubstring(testCases[0], searchString[0]), results[0]);
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```
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`countSubstring("ababababab", "abab")`应返回`2` 。
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```js
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assert.equal(countSubstring(testCases[1], searchString[1]), results[1]);
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```
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`countSubstring("abaabba*bbaba*bbab", "a*b")`应返回`2` 。
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```js
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assert.equal(countSubstring(testCases[2], searchString[2]), results[2]);
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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 testCases = ['the three truths', 'ababababab', 'abaabba*bbaba*bbab'];
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const searchString = ['th', 'abab', 'a*b'];
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const results = [3, 2, 2];
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```
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## --seed-contents--
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```js
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function countSubstring(str, subStr) {
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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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function countSubstring(str, subStr) {
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const escapedSubStr = subStr.replace(/[.+*?^$[\]{}()|/]/g, '\\$&');
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const matches = str.match(new RegExp(escapedSubStr, 'g'));
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return matches ? matches.length : 0;
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}
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```
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