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{"payload":{"allShortcutsEnabled":false,"path":"006. Reverse Integer","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":"006. Reverse Integer/README.md","contentType":"file"},{"name":"main.cpp","path":"006. Reverse Integer/main.cpp","contentType":"file"},{"name":"solution.h","path":"006. Reverse Integer/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这道题也偏容易, 但要注意题目提示的那几种特殊情况. 这体现了思维是否严谨.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e整数反转, 思路很简单, 原数的个位 ==\u0026gt; 结果的最高位. 原数的最高位 ==\u0026gt; 结果的个位. 那么循环一下吧?\u003c/p\u003e\n\u003cdiv class=\"highlight highlight-source-c++ notranslate position-relative overflow-auto\" dir=\"auto\" data-snippet-clipboard-copy-content=\"while (x)\n{\n res = res * 10 + x % 10; // 加号前面是先取得的值保持最高位, 后面则是取得原数的个位.\n x /= 10; // 裁剪原数, 确保能够从低位到高位的值, 能够被依次取得.\n}\"\u003e\u003cpre\u003e\u003cspan class=\"pl-k\"\u003ewhile\u003c/span\u003e (x)\n{\n res = res * \u003cspan class=\"pl-c1\"\u003e10\u003c/span\u003e + x % \u003cspan class=\"pl-c1\"\u003e10\u003c/span\u003e; \u003cspan class=\"pl-c\"\u003e\u003cspan class=\"pl-c\"\u003e//\u003c/span\u003e 加号前面是先取得的值保持最高位, 后面则是取得原数的个位.\u003c/span\u003e\n x /= \u003cspan class=\"pl-c1\"\u003e10\u003c/span\u003e; \u003cspan class=\"pl-c\"\u003e\u003cspan class=\"pl-c\"\u003e//\u003c/span\u003e 裁剪原数, 确保能够从低位到高位的值, 能够被依次取得.\u003c/span\u003e\n}\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e这就是这道题的核心了.\u003c/p\u003e\n\u003chr\u003e\n\u003cp dir=\"auto\"\u003e特殊情况:\u003c/p\u003e\n\u003col dir=\"auto\"\u003e\n\u003cli\u003e后几位是0的情况(10, 100.)? 这个我们的循环已经考虑了, 如果原数的个位是0, res 则会一直保持是0, 直到非0的出现.\u003c/li\u003e\n\u003cli\u003eoverflow的问题? 检测很简单, 将 res 设为 \u003ccode\u003elong\u003c/code\u003e, 那你一个 \u003ccode\u003eint\u003c/code\u003e 怎么反转都不会超出 \u003ccode\u003elong\u003c/code\u003e 的范围了吧? 然后判断 res 和 \u003ccode\u003eINT_MAX\u003c/code\u003e的关系就可以了.\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp dir=\"auto\"\u003e缩减一下上面的循环. 答案有了.\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%2F006.%2520Reverse%2520Integer"}},"totalCount":3,"showBranchInfobar":true},"fileTree":{"":{"items":[{"name":"000. 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