{
   "question_id": 38,
   "question_title": "外观数列",
   "difficulty": "中等",
   "question_content": "<div class=\"notranslate\">\n    <p>给定一个正整数 <code>n</code> ，输出外观数列的第 <code>n</code> 项。</p>\n\n    <p>「外观数列」是一个整数序列，从数字 1 开始，序列中的每一项都是对前一项的描述。</p>\n\n    <p>你可以将其视作是由递归公式定义的数字字符串序列：</p>\n\n    <ul>\n        <li><code>countAndSay(1) = \"1\"</code></li>\n        <li><code>countAndSay(n)</code> 是对 <code>countAndSay(n-1)</code> 的描述，然后转换成另一个数字字符串。</li>\n    </ul>\n\n    <p>前五项如下：</p>\n\n    <pre>\n    1.     1\n    2.     11\n    3.     21\n    4.     1211\n    5.     111221\n    第一项是数字 1 \n    描述前一项，这个数是 1 即 “ 一 个 1 ”，记作 \"11\"\n    描述前一项，这个数是 11 即 “ 二 个 1 ” ，记作 \"21\"\n    描述前一项，这个数是 21 即 “ 一 个 2 + 一 个 1 ” ，记作 \"1211\"\n    描述前一项，这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ，记作 \"111221\"\n    </pre>\n\n    <p>要 <strong>描述</strong> 一个数字字符串，首先要将字符串分割为 <strong>最小</strong> 数量的组，每个组都由连续的最多 <strong>相同字符</strong>\n        组成。然后对于每个组，先描述字符的数量，然后描述字符，形成一个描述组。要将描述转换为数字字符串，先将每组中的字符数量用数字替换，再将所有描述组连接起来。</p>\n\n    <p>例如，数字字符串 <code>\"3322251\"</code> 的描述如下图：</p>\n    <img alt=\"\"\n        src=\"https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0038.Count%20and%20Say/images/countandsay.jpg\"\n        style=\"width: 581px; height: 172px;\" />\n    <ul>\n    </ul>\n\n    <p>&nbsp;</p>\n\n    <p><strong>示例 1：</strong></p>\n\n    <pre><strong>输入：</strong>n = 1\n<strong>输出：</strong>\"1\"\n<strong>解释：</strong>这是一个基本样例。\n</pre>\n\n    <p><strong>示例 2：</strong></p>\n\n    <pre><strong>输入：</strong>n = 4\n<strong>输出：</strong>\"1211\"\n<strong>解释：</strong>\ncountAndSay(1) = \"1\"\ncountAndSay(2) = 读 \"1\" = 一 个 1 = \"11\"\ncountAndSay(3) = 读 \"11\" = 二 个 1 = \"21\"\ncountAndSay(4) = 读 \"21\" = 一 个 2 + 一 个 1 = \"12\" + \"11\" = \"1211\"\n</pre>\n\n    <p>&nbsp;</p>\n\n    <p><strong>提示：</strong></p>\n\n    <ul>\n        <li><code>1 &lt;= n &lt;= 30</code></li>\n    </ul>\n</div>",
   "topic_link": "https://bbs.csdn.net/topics/600470795",
   "cpp": "#include <stdio.h>\n#include <stdlib.h>\n#include <string.h>\nstatic void parse(char *input, char *output)\n{\n\tchar *p = input;\n\tchar *q = output;\n\twhile (*p != '\\0')\n\t{\n\t\tint count = 1;\n\t\twhile (p[0] == p[1])\n\t\t{\n\t\t\tcount++;\n\t\t\tp++;\n\t\t}\n\t\tint n = 0;\n\t\twhile (count > 0)\n\t\t{\n\t\t\tn += count % 10;\n\t\t\tcount /= 10;\n\t\t}\n\t\twhile (n > 0)\n\t\t{\n\t\t\t*q++ = (n % 10) + '0';\n\t\t\tn /= 10;\n\t\t}\n\t\t*q++ = p[0];\n\t\tp++;\n\t}\n\t*q = '\\0';\n}\nstatic char *countAndSay(int n)\n{\n\tif (n < 1)\n\t{\n\t\treturn NULL;\n\t}\n\tchar *result;\n\tchar *prev = malloc(10000);\n\tchar *next = malloc(10000);\n\tstrcpy(prev, \"1\");\n\tif (n == 1)\n\t{\n\t\treturn prev;\n\t}\n\tint i;\n\tfor (i = 2; i <= n; i++)\n\t{\n\t\tif (i & 0x1)\n\t\t{\n\t\t\tparse(next, prev);\n\t\t\tresult = prev;\n\t\t}\n\t\telse\n\t\t{\n\t\t\tparse(prev, next);\n\t\t\tresult = next;\n\t\t}\n\t}\n\treturn result;\n}\nint main(int argc, char **argv)\n{\n\tif (argc != 2)\n\t{\n\t\tfprintf(stderr, \"Usage: ./test n\\n\");\n\t\texit(-1);\n\t}\n\tprintf(\"%s\\n\", countAndSay(atoi(argv[1])));\n\treturn 0;\n}",
   "java": "class Solution {\n\tpublic String countAndSay(int n) {\n\t\tString pre = \"1\";\n\t\tfor (int i = 1; i < n; i++) {\n\t\t\tStringBuilder temp = new StringBuilder();\n\t\t\tchar c = pre.charAt(0);\n\t\t\tint cnt = 1;\n\t\t\tfor (int j = 1; j < pre.length(); j++) {\n\t\t\t\tchar cc = pre.charAt(j);\n\t\t\t\tif (c == cc) {\n\t\t\t\t\tcnt++;\n\t\t\t\t} else {\n\t\t\t\t\ttemp.append(cnt).append(c);\n\t\t\t\t\tcnt = 1;\n\t\t\t\t\tc = cc;\n\t\t\t\t}\n\t\t\t}\n\t\t\ttemp.append(cnt).append(c);\n\t\t\tpre = temp.toString();\n\t\t}\n\t\treturn pre;\n\t}\n}",
   "js": "",
   "python": "class Solution:\n\tdef countAndSay(self, n):\n\t\tif n == 1:\n\t\t\treturn '1'\n\t\tx = '1'\n\t\twhile n > 1:\n\t\t\tx = self.count(x)\n\t\t\tn -= 1\n\t\treturn x\n\tdef count(self, x):\n\t\tm = list(x)\n\t\tres = []\n\t\tm.append(None)\n\t\ti , j = 0 , 0\n\t\twhile i < len(m) - 1:\n\t\t\tj += 1\n\t\t\tif m[j] != m[i]:\n\t\t\t\tres += [j - i, m[i]]\n\t\t\t\ti = j\n\t\treturn ''.join(str(s) for s in res)\n# %%\ns = Solution()\nprint(s.countAndSay(n = 4))",
   "status": 1,
   "keywords": "字符串",
   "license": {
      "cpp": "https://github.com/begeekmyfriend/leetcode",
      "python": "https://github.com/qiyuangong/leetcode",
      "java": "https://blog.csdn.net/a1439775520/article/details/104314641"
   },
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