2.json

{
   "question_id": 3,
   "question_title": "无重复字符的最长子串",
   "difficulty": "中等",
   "question_content": "<p>给定一个字符串，请你找出其中不含有重复字符的 <strong>最长子串 </strong>的长度。</p><p> </p><p><strong>示例 1:</strong></p><pre><strong>输入: </strong>s = \"abcabcbb\"<strong><br />输出: </strong>3 <strong><br />解释:</strong> 因为无重复字符的最长子串是 \"abc\"，所以其长度为 3。</pre><p><strong>示例 2:</strong></p><pre><strong>输入: </strong>s = \"bbbbb\"<strong><br />输出: </strong>1<strong><br />解释: </strong>因为无重复字符的最长子串是 \"b\"，所以其长度为 1。</pre><p><strong>示例 3:</strong></p><pre><strong>输入: </strong>s = \"pwwkew\"<strong><br />输出: </strong>3<strong><br />解释: </strong>因为无重复字符的最长子串是 \"wke\"，所以其长度为 3。 \n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。</pre><p><strong>示例 4:</strong></p><pre><strong>输入: </strong>s = \"\"<strong><br />输出: </strong>0</pre><p> </p><p><strong>提示：</strong></p><ul>\t<li><code>0 <= s.length <= 5 * 10<sup>4</sup></code></li>\t<li><code>s</code> 由英文字母、数字、符号和空格组成</li></ul>",
   "topic_link": "https://bbs.csdn.net/topics/600471011",
   "cpp": "int hset[128];\nint lengthOfLongestSubstring(char *s)\n{\n\tint i = 0, j = 0;\n\tint m = 0;\n\tmemset(hset, 0, sizeof hset);\n\tfor (; s[j]; j++)\n\t{\n\t\ti = hset[s[j]] > i ? hset[s[j]] : i;\n\t\tm = m > j - i + 1 ? m : j - i + 1;\n\t\thset[s[j]] = j + 1;\n\t}\n\treturn m;\n}",
   "java": "class Solution {\n\tpublic int lengthOfLongestSubstring(String s) {\n\t\tint i = 0;\n\t\tint j = 0;\n\t\tint m = 0;\n\t\tMap<Integer, Integer> hset = new HashMap<>();\n\t\tfor (; j < s.length(); j++) {\n\t\t\tint code = s.codePointAt(j);\n\t\t\tInteger o = hset.get(code);\n\t\t\tif (o != null && o > i) {\n\t\t\t\ti = o;\n\t\t\t}\n\t\t\tm = m > j - i + 1 ? m : j - i + 1;\n\t\t\thset.put(code, j + 1);\n\t\t}\n\t\treturn m;\n\t}\n}",
   "js": "\n/**\n * @param {string} s\n * @return {number}\n */\nvar lengthOfLongestSubstring = function(s) {\n    let i = 0;\n    let j = 0;\n    let m = 0;\n    let hset = {};\n    for (;s[j];j++) {\n        i = hset[s[j]] > i ? hset[s[j]] : i;\n        m = m > j - i + 1 ? m : j - i + 1;\n        hset[s[j]] = j + 1;\n    }\n    return m;\n};\n",
   "python": "class Solution:\n\tdef lengthOfLongestSubstring(self, s: str) -> int:\n\t\ti = 0\n\t\tj = 0\n\t\tm = 0\n\t\thset = {}\n\t\twhile j < len(s):\n\t\t\tchar = s[j]\n\t\t\tindex = hset.get(char)\n\t\t\tif index is not None and index > i:\n\t\t\t\ti = index\n\t\t\tm = m if m > j - i + 1 else j - i + 1\n\t\t\thset[char] = j + 1\n\t\t\tj += 1\n\t\treturn m\n# %%\ns = Solution()\nprint(s.lengthOfLongestSubstring('abcabcbb'))",
   "status": 1,
   "keywords": "哈希表,字符串,滑动窗口",
   "license": {
      "cpp": "csdn.net",
      "python": "csdn.net",
      "java": "csdn.net"
   },
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