{"payload":{"allShortcutsEnabled":false,"path":"086. Scramble String","repo":{"id":490101071,"defaultBranch":"master","name":"LeetCode","ownerLogin":"algorithms-forks","currentUserCanPush":false,"isFork":true,"isEmpty":false,"createdAt":"2022-05-09T01:47:36.000Z","ownerAvatar":"https://avatars.githubusercontent.com/u/105182575?v=4","public":true,"private":false,"isOrgOwned":true},"currentUser":null,"refInfo":{"name":"master","listCacheKey":"v0:1652060858.178429","canEdit":false,"refType":"branch","currentOid":"1756256d7e619164076bbf358c8f7ca68cd8bd79"},"tree":{"items":[{"name":"README.md","path":"086. Scramble String/README.md","contentType":"file"},{"name":"TEST.cc","path":"086. Scramble String/TEST.cc","contentType":"file"},{"name":"solution.h","path":"086. Scramble String/solution.h","contentType":"file"}],"templateDirectorySuggestionUrl":null,"readme":{"displayName":"README.md","richText":"\u003carticle class=\"markdown-body entry-content container-lg\" itemprop=\"text\"\u003e\u003cp dir=\"auto\"\u003e两个字符串 s1, s2, 问是否是 scramble. --\u0026gt; 定义问题的解为：\u003ccode\u003eS(s1, s2)\u003c/code\u003e, s1,s2 的长度为 n (若不等，直接 \u003ccode\u003ereturn false;\u003c/code\u003e).\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e然后看给出的例子：\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\" gr|eat rg|tae ----- i == 2 (1)\n / \\ / \\\n g|r e|at r|g ta|e ----- i == 1 (2)\n / \\ / \\ / \\ / \\\n g r e a|t r g t|a e ----- i == 1 (3)\n / \\ / \\\n a t t a\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003e gr|eat rg|tae ----- i == 2 (1)\n / \\ / \\\n g|r e|at r|g ta|e ----- i == 1 (2)\n / \\ / \\ / \\ / \\\n g r e a|t r g t|a e ----- i == 1 (3)\n / \\ / \\\n a t t a\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e可以来看看，这里面究竟有几种情况。按照上图所示的步骤进行分析：\u003c/p\u003e\n\u003col dir=\"auto\"\u003e\n\u003cli\u003e\u003ccode\u003eS(s1, s2) = S(s1.substr(0,i), s2.substr(0,i)) \u0026amp;\u0026amp; S(s1.substr(i), s2.substr(i))\u003c/code\u003e (第一种可能)\u003c/li\u003e\n\u003cli\u003e重点看 \u003ccode\u003ee|at\u003c/code\u003e 与 \u003ccode\u003eta|e\u003c/code\u003e, 可以推得: \u003ccode\u003eS(s1, s2) = S(s1.substr(0,i), s2.substr(n-i)) \u0026amp;\u0026amp; S(s1.substr(i), s2.substr(0, n-i))\u003c/code\u003e (第二种可能)\u003c/li\u003e\n\u003cli\u003e可以并入 2. 但可以看出 S 最简单的解: \u003ccode\u003eS(\"a\", \"a\") == true\u003c/code\u003e 以及 \u003ccode\u003eS(\"t\", \"t\") == true\u003c/code\u003e. 即 \u003ccode\u003eif (s1 == s2) return true;\u003c/code\u003e\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp dir=\"auto\"\u003e综上三步，可以得到递归的公式：\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\" _________ true; \n S(s1, s2) _|s1 == s2\n | __ S(s1.substr(0,i), s2.substr(0,i)) \u0026amp;\u0026amp; S(s1.substr(i), s2.substr(i))\n |_ or |\n |__ S(s1.substr(0,i), s2.substr(n-i)) \u0026amp;\u0026amp; S(s1.substr(i), s2.substr(n-i))\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003e _________ true; \n S(s1, s2) _|s1 == s2\n | __ S(s1.substr(0,i), s2.substr(0,i)) \u0026amp;\u0026amp; S(s1.substr(i), s2.substr(i))\n |_ or |\n |__ S(s1.substr(0,i), s2.substr(n-i)) \u0026amp;\u0026amp; S(s1.substr(i), s2.substr(n-i))\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e显然 i 的范围是 [1 ,n-1].\u003c/p\u003e\n\u003chr\u003e\n\u003cp dir=\"auto\"\u003e有了这个公式做支撑，我们基本看到了递归的雏形。也可以试着用这个来写出代码。\u003c/p\u003e\n\u003cdiv class=\"highlight highlight-source-c++ notranslate position-relative overflow-auto\" dir=\"auto\" data-snippet-clipboard-copy-content=\"if (s1.empty() || s2.empty() || s1.size() != s2.size()) return false;\nelse if (s1 == s2) return true;\nfor (size_t i=1; i\u0026lt;s1.size(); ++i)\n if (isScramble(s1.substr(0,i), s2.substr(0,i)) \u0026amp;\u0026amp; isScramble(s1.substr(i), s2.substr(i))) return true;\n else if (isScramble(s1.substr(0,i), s2.substr(s2.size()-i)) \u0026amp;\u0026amp; isScramble(s1.substr(i), s2.substr(0, s2.size()-i))) return true;\nreturn false;\"\u003e\u003cpre\u003e\u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (s1.empty() || s2.empty() || s1.size() != s2.size()) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003efalse\u003c/span\u003e;\n\u003cspan class=\"pl-k\"\u003eelse\u003c/span\u003e \u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (s1 == s2) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003etrue\u003c/span\u003e;\n\u003cspan class=\"pl-k\"\u003efor\u003c/span\u003e (\u003cspan class=\"pl-c1\"\u003esize_t\u003c/span\u003e i=\u003cspan class=\"pl-c1\"\u003e1\u003c/span\u003e; i\u0026lt;s1.size(); ++i)\n \u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (isScramble(s1.substr(\u003cspan class=\"pl-c1\"\u003e0\u003c/span\u003e,i), s2.substr(\u003cspan class=\"pl-c1\"\u003e0\u003c/span\u003e,i)) \u0026amp;\u0026amp; isScramble(s1.substr(i), s2.substr(i))) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003etrue\u003c/span\u003e;\n \u003cspan class=\"pl-k\"\u003eelse\u003c/span\u003e \u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (isScramble(s1.substr(\u003cspan class=\"pl-c1\"\u003e0\u003c/span\u003e,i), s2.substr(s2.size()-i)) \u0026amp;\u0026amp; isScramble(s1.substr(i), s2.substr(\u003cspan class=\"pl-c1\"\u003e0\u003c/span\u003e, s2.size()-i))) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003etrue\u003c/span\u003e;\n\u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003efalse\u003c/span\u003e;\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e六行代码基本已经从逻辑上解出了这道题。但如果去 LeetCode 上提交，会发现 TimeOut. 因为这样其实与枚举无异。如果遇到连字符都不尽相同的两个字符串，实在是浪费时间。\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e于是我们考虑排除两种情况：\u003c/p\u003e\n\u003col dir=\"auto\"\u003e\n\u003cli\u003es1 与 s2 中字符组成有差异，如 s1 中有 'g', s2 中却无。直接 \u003ccode\u003ereturn false;\u003c/code\u003e\u003c/li\u003e\n\u003cli\u003es1 与 s2 中字符出现的个数有差异，如 s1 中 'g' 出现两次，而 s2 中却只有一次。直接 \u003ccode\u003ereturn false;\u003c/code\u003e.\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp dir=\"auto\"\u003e根据这俩个条件，我们在第三行补充代码：\u003c/p\u003e\n\u003cdiv class=\"highlight highlight-source-c++ notranslate position-relative overflow-auto\" dir=\"auto\" data-snippet-clipboard-copy-content=\"for(auto c : s1)\n if (s2.find_first_of(c) == string::npos) return false;\n else if (std::count(s1.cbegin(), s1.cend(), c) != std::count(s2.cbegin(), s2.cend(), c)) return false;\"\u003e\u003cpre\u003e\u003cspan class=\"pl-k\"\u003efor\u003c/span\u003e(\u003cspan class=\"pl-k\"\u003eauto\u003c/span\u003e c : s1)\n \u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (s2.find_first_of(c) == string::npos) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003efalse\u003c/span\u003e;\n \u003cspan class=\"pl-k\"\u003eelse\u003c/span\u003e \u003cspan class=\"pl-k\"\u003eif\u003c/span\u003e (std::count(s1.cbegin(), s1.cend(), c) != std::count(s2.cbegin(), s2.cend(), c)) \u003cspan class=\"pl-k\"\u003ereturn\u003c/span\u003e \u003cspan class=\"pl-c1\"\u003efalse\u003c/span\u003e;\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e然后提交，AC, 16ms. 问题顺利解决。\u003c/p\u003e\n\u003c/article\u003e","errorMessage":null,"headerInfo":{"toc":[],"siteNavLoginPath":"/login?return_to=https%3A%2F%2Fgithub.com%2Falgorithms-forks%2FLeetCode%2Ftree%2Fmaster%2F086.%2520Scramble%2520String"}},"totalCount":3,"showBranchInfobar":true},"fileTree":{"":{"items":[{"name":"000. 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Generate Parentheses","contentType":"directory"},{"name":"022. Merge k Sorted Lists","path":"022. Merge k Sorted Lists","contentType":"directory"},{"name":"023. Swap Nodes in Pairs","path":"023. Swap Nodes in Pairs","contentType":"directory"},{"name":"024. Reverse Nodes in k-Group","path":"024. Reverse Nodes in k-Group","contentType":"directory"},{"name":"025. Remove Duplicates from Sorted Array","path":"025. Remove Duplicates from Sorted Array","contentType":"directory"},{"name":"026. Remove Element","path":"026. Remove Element","contentType":"directory"},{"name":"027. Implement strStr()","path":"027. Implement strStr()","contentType":"directory"},{"name":"030. Next Permutation","path":"030. Next Permutation","contentType":"directory"},{"name":"032. Search in Rotated Sorted Array","path":"032. Search in Rotated Sorted Array","contentType":"directory"},{"name":"033. Search for a Range","path":"033. Search for a Range","contentType":"directory"},{"name":"034. 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