{"payload":{"allShortcutsEnabled":false,"path":"127. Longest Consecutive Sequence","repo":{"id":189527355,"defaultBranch":"master","name":"LeetCode","ownerLogin":"MyDataSource","currentUserCanPush":false,"isFork":true,"isEmpty":false,"createdAt":"2019-05-31T04:26:14.000Z","ownerAvatar":"https://avatars.githubusercontent.com/u/23212692?v=4","public":true,"private":false,"isOrgOwned":true},"currentUser":null,"refInfo":{"name":"master","listCacheKey":"v0:1639597980.8342059","canEdit":false,"refType":"branch","currentOid":"1756256d7e619164076bbf358c8f7ca68cd8bd79"},"tree":{"items":[{"name":"README.md","path":"127. Longest Consecutive Sequence/README.md","contentType":"file"},{"name":"TEST.cc","path":"127. Longest Consecutive Sequence/TEST.cc","contentType":"file"},{"name":"solution.h","path":"127. Longest Consecutive Sequence/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这道题定为 hard 级别的原因并非是解决思路有多难想, 而是难在了效率的瓶颈.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e要求在 O(n) 的复杂度情况下解决. 要知道, 完整的迭代一次就需要 O(n), 这相当于要求在一次迭代过程中, 计算出最大连续序列的长度来.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e比较直接的思路是, 先排个序, 然后完整迭代, 中途记录 max 连击. 这个思路非常简单, 但且不论排序, 那一次完整的迭代就几乎要超标.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e这种情况下, 以空间换时间就是最佳的选择. 如果我们在迭代过程中, 尝试记录当前最大的连续序列长度呢? 这几乎是矛盾的, 因为原数组并非有序. 如何可以知道连续序列的存在?\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e这让我想到了生活中一个真实的用例: 整理书籍. 学生时代, 每当大考之后, 都会清理书籍, 入库或者变卖. 我们是如何分类清理的呢? 遇到一本书, 看其属于哪一类, 然后放在一个固定的位置. 当全部书籍翻过一遍之后, 就会很自然的出现一堆堆的书堆. 每一堆都是同一类书籍. 我们很清楚的看到哪一类的书籍最多. 最高的那个就是啦.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e这里也是一样, 如果我们弄一个 hash 表(key 为数组值, value 为连续序列 size), 将数组里每一个元素都放入表中, 然后发现连续, 则进行 \u003ccode\u003e+1\u003c/code\u003e 操作. 具体来说, 就是:\u003c/p\u003e\n\u003cul dir=\"auto\"\u003e\n\u003cli\u003e若 map[i+1] 和 map[i-1] 存在, 那么 i-1, i, i+1 构成一个三元序列, size == 3;\u003c/li\u003e\n\u003cli\u003e若只有 map[i+1] 或者只有 map[i-1], 那么 size = 2;\u003c/li\u003e\n\u003cli\u003e若 map[i+1] 和 map[i-1] 都不存在, 那么 size = 1; (自己)\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp dir=\"auto\"\u003e上面三种情况可以写成一个式子:\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\"map[i] = map[i-1] + map[i+1] + 1; // 左边记录的序列长度 + 右边记录的序列长度 + 自己\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003emap[i] = map[i-1] + map[i+1] + 1; // 左边记录的序列长度 + 右边记录的序列长度 + 自己\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e但当 map[i] 的值更新后, 我们还应该将其最新值(长度)扩展至该连续序列的边界上去. 可是如何扩展呢?\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e如 map[i] 的左边最远的连续节点是谁? 右边的最远连续节点是谁? 显然左边可以递推至 map[i-1] 那去, 右边同理. map[i-1] 如果等于1, 表示到它这就断了, 如果等于2, 那么表示它的左边还有一个连续节点.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e那么这个节点的位置也可以用一个式子表示:\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\"map[i - map[i-1]] // 左边最远节点, 用当前位置 - 左边的长度, 正好获得最左连续节点的坐标.\nmap[i + map[i+1]] // 右边同理.\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003emap[i - map[i-1]] // 左边最远节点, 用当前位置 - 左边的长度, 正好获得最左连续节点的坐标.\nmap[i + map[i+1]] // 右边同理.\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e所以当 map[i] 的值更新后, 应推广至最左与最右. 即:\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\"map[i - map[i-1]] = map[i + map[i+1]] = map[i] = map[i-1] + map[i+1] + 1;\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003emap[i - map[i-1]] = map[i + map[i+1]] = map[i] = map[i-1] + map[i+1] + 1;\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e最后, 由于我们只关心最长连续序列的长度, 所以我们设置一个变量来记录它, 如 max.\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\"max = std::max(max, map[i]);\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003emax = std::max(max, map[i]);\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e综合上述几个式子, 可以归并为一句话:\u003c/p\u003e\n\u003cdiv class=\"snippet-clipboard-content notranslate position-relative overflow-auto\" data-snippet-clipboard-copy-content=\"max = std::max(max, map[i] = map[i - map[i-1]] = map[i + map[i+1]] = map[i-1] + map[i+1] + 1);\"\u003e\u003cpre class=\"notranslate\"\u003e\u003ccode\u003emax = std::max(max, map[i] = map[i - map[i-1]] = map[i + map[i+1]] = map[i-1] + map[i+1] + 1);\n\u003c/code\u003e\u003c/pre\u003e\u003c/div\u003e\n\u003cp dir=\"auto\"\u003e这句话即为核心了.\u003c/p\u003e\n\u003cp dir=\"auto\"\u003e由于哈希表 unordered_map 默认值为 0. 所以如果遇到一个新值, 则记 map[i] == 1. 如果 map[i] != 0, 显然这个值我们已经存在于 hash 表中, 可以舍弃, 进行下一次迭代.\u003c/p\u003e\n\u003c/article\u003e","errorMessage":null,"headerInfo":{"toc":[],"siteNavLoginPath":"/login?return_to=https%3A%2F%2Fgithub.com%2FMyDataSource%2FLeetCode%2Ftree%2Fmaster%2F127.%2520Longest%2520Consecutive%2520Sequence"}},"totalCount":3,"showBranchInfobar":true},"fileTree":{"":{"items":[{"name":"000. 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