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style: update description to lc problem: No.0003
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solution/0000-0099/0003.Longest Substring Without Repeating Characters/README.md

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<pre>
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<strong>输入: </strong>s = "abcabcbb"
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<strong>输出: </strong>3
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<strong>解释:</strong> 因为无重复字符的最长子串是 <code>"abc",所以其</code>长度为 3。
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<strong>解释:</strong> 因为无重复字符的最长子串是 "abc",所以其长度为 3。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入: </strong>s = "bbbbb"
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<strong>输出: </strong>1
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<strong>解释: </strong>因为无重复字符的最长子串是 <code>"b"</code>,所以其长度为 1。
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<strong>解释: </strong>因为无重复字符的最长子串是 "b",所以其长度为 1。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入: </strong>s = "pwwkew"
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<strong>输出: </strong>3
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<strong>解释: </strong>因为无重复字符的最长子串是&nbsp;<code>"wke"</code>,所以其长度为 3。
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&nbsp; 请注意,你的答案必须是 <strong>子串 </strong>的长度,<code>"pwke"</code>&nbsp;是一个<em>子序列,</em>不是子串。
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<strong>解释: </strong>因为无重复字符的最长子串是 "wke",所以其长度为 3。
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请注意,你的答案必须是 <strong>子串 </strong>的长度,"pwke" 是一个<em>子序列,</em>不是子串。
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</pre>
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<p>&nbsp;</p>

solution/main.py

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def refresh(result):
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"""update problems"""
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pattern = re.compile("src=\"(.*?)\"")
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skip_question_ids = {1599}
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skip_question_ids = {3, 1599}
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for question in result:
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front_question_id = question['frontend_question_id']

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