mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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188 lines
25 KiB
JSON
188 lines
25 KiB
JSON
{
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"title": "Accounts Merge",
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"content": "<p>Given a list of <code>accounts</code> where each element <code>accounts[i]</code> is a list of strings, where the first element <code>accounts[i][0]</code> is a name, and the rest of the elements are <strong>emails</strong> representing emails of the account.</p>\n\n<p>Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.</p>\n\n<p>After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails <strong>in sorted order</strong>. The accounts themselves can be returned in <strong>any order</strong>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> accounts = [["John","johnsmith@mail.com","john_newyork@mail.com"],["John","johnsmith@mail.com","john00@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]]\n<strong>Output:</strong> [["John","john00@mail.com","john_newyork@mail.com","johnsmith@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]]\n<strong>Explanation:</strong>\nThe first and second John's are the same person as they have the common email "johnsmith@mail.com".\nThe third John and Mary are different people as none of their email addresses are used by other accounts.\nWe could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], \n['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> accounts = [["Gabe","Gabe0@m.co","Gabe3@m.co","Gabe1@m.co"],["Kevin","Kevin3@m.co","Kevin5@m.co","Kevin0@m.co"],["Ethan","Ethan5@m.co","Ethan4@m.co","Ethan0@m.co"],["Hanzo","Hanzo3@m.co","Hanzo1@m.co","Hanzo0@m.co"],["Fern","Fern5@m.co","Fern1@m.co","Fern0@m.co"]]\n<strong>Output:</strong> [["Ethan","Ethan0@m.co","Ethan4@m.co","Ethan5@m.co"],["Gabe","Gabe0@m.co","Gabe1@m.co","Gabe3@m.co"],["Hanzo","Hanzo0@m.co","Hanzo1@m.co","Hanzo3@m.co"],["Kevin","Kevin0@m.co","Kevin3@m.co","Kevin5@m.co"],["Fern","Fern0@m.co","Fern1@m.co","Fern5@m.co"]]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= accounts.length <= 1000</code></li>\n\t<li><code>2 <= accounts[i].length <= 10</code></li>\n\t<li><code>1 <= accounts[i][j] <= 30</code></li>\n\t<li><code>accounts[i][0]</code> consists of English letters.</li>\n\t<li><code>accounts[i][j] (for j > 0)</code> is a valid email.</li>\n</ul>\n",
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"translatedTitle": "账户合并",
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"translatedContent": "<p>给定一个列表 <code>accounts</code>，每个元素 <code>accounts[i]</code> 是一个字符串列表，其中第一个元素 <code>accounts[i][0]</code> 是 <em>名称 (name)</em>，其余元素是 <em><strong>emails</strong> </em>表示该账户的邮箱地址。</p>\n\n<p>现在，我们想合并这些账户。如果两个账户都有一些共同的邮箱地址，则两个账户必定属于同一个人。请注意，即使两个账户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的账户，但其所有账户都具有相同的名称。</p>\n\n<p>合并账户后，按以下格式返回账户：每个账户的第一个元素是名称，其余元素是 <strong>按字符 ASCII 顺序排列</strong> 的邮箱地址。账户本身可以以 <strong>任意顺序</strong> 返回。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>accounts = [[\"John\", \"johnsmith@mail.com\", \"john00@mail.com\"], [\"John\", \"johnnybravo@mail.com\"], [\"John\", \"johnsmith@mail.com\", \"john_newyork@mail.com\"], [\"Mary\", \"mary@mail.com\"]]\n<b>输出：</b>[[\"John\", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'], [\"John\", \"johnnybravo@mail.com\"], [\"Mary\", \"mary@mail.com\"]]\n<b>解释：</b>\n第一个和第三个 John 是同一个人，因为他们有共同的邮箱地址 \"johnsmith@mail.com\"。 \n第二个 John 和 Mary 是不同的人，因为他们的邮箱地址没有被其他帐户使用。\n可以以任何顺序返回这些列表，例如答案 [['Mary'，'mary@mail.com']，['John'，'johnnybravo@mail.com']，\n['John'，'john00@mail.com'，'john_newyork@mail.com'，'johnsmith@mail.com']] 也是正确的。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>accounts = [[\"Gabe\",\"Gabe0@m.co\",\"Gabe3@m.co\",\"Gabe1@m.co\"],[\"Kevin\",\"Kevin3@m.co\",\"Kevin5@m.co\",\"Kevin0@m.co\"],[\"Ethan\",\"Ethan5@m.co\",\"Ethan4@m.co\",\"Ethan0@m.co\"],[\"Hanzo\",\"Hanzo3@m.co\",\"Hanzo1@m.co\",\"Hanzo0@m.co\"],[\"Fern\",\"Fern5@m.co\",\"Fern1@m.co\",\"Fern0@m.co\"]]\n<strong>输出：</strong>[[\"Ethan\",\"Ethan0@m.co\",\"Ethan4@m.co\",\"Ethan5@m.co\"],[\"Gabe\",\"Gabe0@m.co\",\"Gabe1@m.co\",\"Gabe3@m.co\"],[\"Hanzo\",\"Hanzo0@m.co\",\"Hanzo1@m.co\",\"Hanzo3@m.co\"],[\"Kevin\",\"Kevin0@m.co\",\"Kevin3@m.co\",\"Kevin5@m.co\"],[\"Fern\",\"Fern0@m.co\",\"Fern1@m.co\",\"Fern5@m.co\"]]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= accounts.length <= 1000</code></li>\n\t<li><code>2 <= accounts[i].length <= 10</code></li>\n\t<li><code>1 <= accounts[i][j].length <= 30</code></li>\n\t<li><code>accounts[i][0]</code> 由英文字母组成</li>\n\t<li><code>accounts[i][j] (for j > 0)</code> 是有效的邮箱地址</li>\n</ul>\n",
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