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leetcode-problemset/leetcode-cn/originData/repeated-string-match.json
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{
"data": {
"question": {
"questionId": "686",
"questionFrontendId": "686",
"categoryTitle": "Algorithms",
"boundTopicId": 1284,
"title": "Repeated String Match",
"titleSlug": "repeated-string-match",
"content": "<p>Given two strings <code>a</code> and <code>b</code>, return <em>the minimum number of times you should repeat string </em><code>a</code><em> so that string</em> <code>b</code> <em>is a substring of it</em>. If it is impossible for <code>b</code> to be a substring of <code>a</code> after repeating it, return <code>-1</code>.</p>\n\n<p><strong>Notice:</strong> string <code>&quot;abc&quot;</code> repeated 0 times is <code>&quot;&quot;</code>, repeated 1 time is <code>&quot;abc&quot;</code> and repeated 2 times is <code>&quot;abcabc&quot;</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> a = &quot;abcd&quot;, b = &quot;cdabcdab&quot;\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We return 3 because by repeating a three times &quot;ab<strong>cdabcdab</strong>cd&quot;, b is a substring of it.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> a = &quot;a&quot;, b = &quot;aa&quot;\n<strong>Output:</strong> 2\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= a.length, b.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>a</code> and <code>b</code> consist of lowercase English letters.</li>\n</ul>\n",
"translatedTitle": "重复叠加字符串匹配",
"translatedContent": "<p>给定两个字符串&nbsp;<code>a</code> 和 <code>b</code>,寻找重复叠加字符串 <code>a</code> 的最小次数,使得字符串 <code>b</code> 成为叠加后的字符串 <code>a</code> 的子串,如果不存在则返回 <code>-1</code>。</p>\n\n<p><strong>注意:</strong>字符串 <code>&quot;abc&quot;</code>&nbsp;重复叠加 0 次是 <code>&quot;&quot;</code>,重复叠加 1 次是&nbsp;<code>&quot;abc&quot;</code>,重复叠加 2 次是&nbsp;<code>&quot;abcabc&quot;</code>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre><strong>输入:</strong>a = &quot;abcd&quot;, b = &quot;cdabcdab&quot;\n<strong>输出:</strong>3\n<strong>解释:</strong>a 重复叠加三遍后为 &quot;ab<strong>cdabcdab</strong>cd&quot;, 此时 b 是其子串。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre><strong>输入:</strong>a = &quot;a&quot;, b = &quot;aa&quot;\n<strong>输出:</strong>2\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre><strong>输入:</strong>a = &quot;a&quot;, b = &quot;a&quot;\n<strong>输出:</strong>1\n</pre>\n\n<p><strong>示例 4</strong></p>\n\n<pre><strong>输入:</strong>a = &quot;abc&quot;, b = &quot;wxyz&quot;\n<strong>输出:</strong>-1\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= a.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= b.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>a</code> 和 <code>b</code> 由小写英文字母组成</li>\n</ul>\n",
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"code": "class Solution(object):\n def repeatedStringMatch(self, a, b):\n \"\"\"\n :type a: str\n :type b: str\n :rtype: int\n \"\"\"",
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