2.1 KiB
2.1 KiB
title | id | challengeType | videoUrl | localeTitle |
---|---|---|---|---|
Count occurrences of a substring | 596fda99c69f779975a1b67d | 5 | 计算子字符串的出现次数 |
Description
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。
该函数应该有两个参数:
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。它应该返回一个整数计数。
匹配应产生最多数量的非重叠匹配。
通常,这实质上意味着从左到右或从右到左匹配。
Instructions
Tests
tests:
- text: <code>countSubstring</code>是一个函数。
testString: 'assert(typeof countSubstring === "function", "<code>countSubstring</code> is a function.");'
- text: '<code>countSubstring("the three truths", "th")</code>应该返回<code>3</code> 。'
testString: 'assert.equal(countSubstring(testCases[0], searchString[0]), results[0], "<code>countSubstring("the three truths", "th")</code> should return <code>3</code>.");'
- text: '<code>countSubstring("ababababab", "abab")</code>应返回<code>2</code> 。'
testString: 'assert.equal(countSubstring(testCases[1], searchString[1]), results[1], "<code>countSubstring("ababababab", "abab")</code> should return <code>2</code>.");'
- text: '<code>countSubstring("abaabba*bbaba*bbab", "a*b")</code>应返回<code>2</code> 。'
testString: 'assert.equal(countSubstring(testCases[2], searchString[2]), results[2], "<code>countSubstring("abaabba*bbaba*bbab", "a*b")</code> should return <code>2</code>.");'
Challenge Seed
function countSubstring (str, subStr) {
// Good luck!
return true;
}
After Test
console.info('after the test');
Solution
// solution required