leetcode-problemset/leetcode-cn/originData/remove-all-occurrences-of-a-substring.json

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        "question": {
            "questionId": "2021",
            "questionFrontendId": "1910",
            "categoryTitle": "Algorithms",
            "boundTopicId": 842644,
            "title": "Remove All Occurrences of a Substring",
            "titleSlug": "remove-all-occurrences-of-a-substring",
            "content": "<p>Given two strings <code>s</code> and <code>part</code>, perform the following operation on <code>s</code> until <strong>all</strong> occurrences of the substring <code>part</code> are removed:</p>\n\n<ul>\n\t<li>Find the <strong>leftmost</strong> occurrence of the substring <code>part</code> and <strong>remove</strong> it from <code>s</code>.</li>\n</ul>\n\n<p>Return <code>s</code><em> after removing all occurrences of </em><code>part</code>.</p>\n\n<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;daabcbaabcbc&quot;, part = &quot;abc&quot;\n<strong>Output:</strong> &quot;dab&quot;\n<strong>Explanation</strong>: The following operations are done:\n- s = &quot;da<strong><u>abc</u></strong>baabcbc&quot;, remove &quot;abc&quot; starting at index 2, so s = &quot;dabaabcbc&quot;.\n- s = &quot;daba<strong><u>abc</u></strong>bc&quot;, remove &quot;abc&quot; starting at index 4, so s = &quot;dababc&quot;.\n- s = &quot;dab<strong><u>abc</u></strong>&quot;, remove &quot;abc&quot; starting at index 3, so s = &quot;dab&quot;.\nNow s has no occurrences of &quot;abc&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;axxxxyyyyb&quot;, part = &quot;xy&quot;\n<strong>Output:</strong> &quot;ab&quot;\n<strong>Explanation</strong>: The following operations are done:\n- s = &quot;axxx<strong><u>xy</u></strong>yyyb&quot;, remove &quot;xy&quot; starting at index 4 so s = &quot;axxxyyyb&quot;.\n- s = &quot;axx<strong><u>xy</u></strong>yyb&quot;, remove &quot;xy&quot; starting at index 3 so s = &quot;axxyyb&quot;.\n- s = &quot;ax<strong><u>xy</u></strong>yb&quot;, remove &quot;xy&quot; starting at index 2 so s = &quot;axyb&quot;.\n- s = &quot;a<strong><u>xy</u></strong>b&quot;, remove &quot;xy&quot; starting at index 1 so s = &quot;ab&quot;.\nNow s has no occurrences of &quot;xy&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= part.length &lt;= 1000</code></li>\n\t<li><code>s</code>​​​​​​ and <code>part</code> consists of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "删除一个字符串中所有出现的给定子字符串",
            "translatedContent": "<p>给你两个字符串 <code>s</code> 和 <code>part</code> ，请你对 <code>s</code> 反复执行以下操作直到 <b>所有</b> 子字符串 <code>part</code> 都被删除：</p>\n\n<ul>\n\t<li>找到 <code>s</code> 中 <strong>最左边</strong> 的子字符串 <code>part</code> ，并将它从 <code>s</code> 中删除。</li>\n</ul>\n\n<p>请你返回从 <code>s</code> 中删除所有 <code>part</code> 子字符串以后得到的剩余字符串。</p>\n\n<p>一个 <strong>子字符串</strong> 是一个字符串中连续的字符序列。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>s = \"daabcbaabcbc\", part = \"abc\"\n<b>输出：</b>\"dab\"\n<b>解释：</b>以下操作按顺序执行：\n- s = \"da<strong>abc</strong>baabcbc\" ，删除下标从 2 开始的 \"abc\" ，得到 s = \"dabaabcbc\" 。\n- s = \"daba<strong>abc</strong>bc\" ，删除下标从 4 开始的 \"abc\" ，得到 s = \"dababc\" 。\n- s = \"dab<strong>abc</strong>\" ，删除下标从 3 开始的 \"abc\" ，得到 s = \"dab\" 。\n此时 s 中不再含有子字符串 \"abc\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>s = \"axxxxyyyyb\", part = \"xy\"\n<b>输出：</b>\"ab\"\n<b>解释：</b>以下操作按顺序执行：\n- s = \"axxx<strong>xy</strong>yyyb\" ，删除下标从 4 开始的 \"xy\" ，得到 s = \"axxxyyyb\" 。\n- s = \"axx<strong>xy</strong>yyb\" ，删除下标从 3 开始的 \"xy\" ，得到 s = \"axxyyb\" 。\n- s = \"ax<strong>xy</strong>yb\" ，删除下标从 2 开始的 \"xy\" ，得到 s = \"axyb\" 。\n- s = \"a<strong>xy</strong>b\" ，删除下标从 1 开始的 \"xy\" ，得到 s = \"ab\" 。\n此时 s 中不再含有子字符串 \"xy\" 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= part.length &lt;= 1000</code></li>\n\t<li><code>s</code>​​​​​​ 和 <code>part</code> 只包小写英文字母。</li>\n</ul>\n",
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                "Note that a new occurrence of pattern can appear if you remove an old one, For example, s = \"ababcc\" and pattern = \"abc\".",
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