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feat: update lc problems (doocs#2280)
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‎solution/0100-0199/0197.Rising Temperature/README.md‎

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| temperature | int |
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+---------------+---------+
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id 是该表具有唯一值的列。
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没有具有相同 recordDate 的不同行。
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该表包含特定日期的温度信息</pre>
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<p>&nbsp;</p>

‎solution/0100-0199/0197.Rising Temperature/README_EN.md‎

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| temperature | int |
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+---------------+---------+
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id is the column with unique values for this table.
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There are no different rows with the same recordDate.
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This table contains information about the temperature on a certain day.
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</pre>
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‎solution/0500-0599/0521.Longest Uncommon Subsequence I/README_EN.md‎

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## Description
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<p>Given two strings <code>a</code> and <code>b</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between </em><code>a</code> <em>and</em> <code>b</code>. If the longest uncommon subsequence does not exist, return <code>-1</code>.</p>
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<p>Given two strings <code>a</code> and <code>b</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between </em><code>a</code> <em>and</em> <code>b</code>. <em>If no such uncommon subsequence exists, return</em> <code>-1</code><em>.</em></p>
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<p>An <strong>uncommon subsequence</strong> between two strings is a string that is a <strong>subsequence of one but not the other</strong>.</p>
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<p>A <strong>subsequence</strong> of a string <code>s</code> is a string that can be obtained after deleting any number of characters from <code>s</code>.</p>
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<ul>
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<li>For example, <code>&quot;abc&quot;</code> is a subsequence of <code>&quot;aebdc&quot;</code> because you can delete the underlined characters in <code>&quot;a<u>e</u>b<u>d</u>c&quot;</code> to get <code>&quot;abc&quot;</code>. Other subsequences of <code>&quot;aebdc&quot;</code> include <code>&quot;aebdc&quot;</code>, <code>&quot;aeb&quot;</code>, and <code>&quot;&quot;</code> (empty string).</li>
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</ul>
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<p>An <strong>uncommon subsequence</strong> between two strings is a string that is a <strong><span data-keyword="subsequence-string">subsequence</span> of exactly one of them</strong>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> a = &quot;aaa&quot;, b = &quot;aaa&quot;
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<strong>Output:</strong> -1
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<strong>Explanation:</strong>&nbsp;Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a.
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<strong>Explanation:</strong>&nbsp;Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a. So the answer would be <code>-1</code>.
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</pre>
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<p>&nbsp;</p>

‎solution/0600-0699/0606.Construct String from Binary Tree/README_EN.md‎

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## Description
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<p>Given the <code>root</code> of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.</p>
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<p>Given the <code>root</code> node of a binary tree, your task is to create a string representation of the tree following a specific set of formatting rules. The representation should be based on a preorder traversal of the binary tree and must adhere to the following guidelines:</p>
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<p>Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.</p>
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<ul>
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<li>
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<p><strong>Node Representation</strong>: Each node in the tree should be represented by its integer value.</p>
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</li>
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<li>
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<p><strong>Parentheses for Children</strong>: If a node has at least one child (either left or right), its children should be represented inside parentheses. Specifically:</p>
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<ul>
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<li>If a node has a left child, the value of the left child should be enclosed in parentheses immediately following the node&#39;s value.</li>
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<li>If a node has a right child, the value of the right child should also be enclosed in parentheses. The parentheses for the right child should follow those of the left child.</li>
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</ul>
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</li>
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<li>
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<p><strong>Omitting Empty Parentheses</strong>: Any empty parentheses pairs (i.e., <code>()</code>) should be omitted from the final string representation of the tree, with one specific exception: when a node has a right child but no left child. In such cases, you must include an empty pair of parentheses to indicate the absence of the left child. This ensures that the one-to-one mapping between the string representation and the original binary tree structure is maintained.</p>
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<p>In summary, empty parentheses pairs should be omitted when a node has only a left child or no children. However, when a node has a right child but no left child, an empty pair of parentheses must precede the representation of the right child to reflect the tree&#39;s structure accurately.</p>
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</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons1-tree.jpg" style="width: 292px; height: 301px;" />
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons1-tree.jpg" style="padding: 10px; background: #fff; border-radius: .5rem;" />
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<pre>
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<strong>Input:</strong> root = [1,2,3,4]
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<strong>Output:</strong> &quot;1(2(4))(3)&quot;
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<strong>Explanation:</strong> Originally, it needs to be &quot;1(2(4)())(3()())&quot;, but you need to omit all the unnecessary empty parenthesis pairs. And it will be &quot;1(2(4))(3)&quot;
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<strong>Explanation:</strong> Originally, it needs to be &quot;1(2(4)())(3()())&quot;, but you need to omit all the empty parenthesis pairs. And it will be &quot;1(2(4))(3)&quot;.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons2-tree.jpg" style="width: 207px; height: 293px;" />
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons2-tree.jpg" style="padding: 10px; background: #fff; border-radius: .5rem;" />
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<pre>
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<strong>Input:</strong> root = [1,2,3,null,4]
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<strong>Output:</strong> &quot;1(2()(4))(3)&quot;
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<strong>Explanation:</strong> Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
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<strong>Explanation:</strong> Almost the same as the first example, except the <code>()</code> after <code>2</code> is necessary to indicate the absence of a left child for <code>2</code> and the presence of a right child.
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</pre>
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<p>&nbsp;</p>

‎solution/0800-0899/0827.Making A Large Island/README.md‎

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<!-- 这里写题目描述 -->
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<p>给你一个大小为 <code>n x n</code> 二进制矩阵 <code>grid</code> 。<strong>最多</strong> 只能将一格<code>0</code> 变成<code>1</code> 。</p>
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<p>给你一个大小为 <code>n x n</code> 二进制矩阵 <code>grid</code> 。<strong>最多</strong> 只能将一格&nbsp;<code>0</code> 变成&nbsp;<code>1</code> 。</p>
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<p>返回执行此操作后,<code>grid</code> 中最大的岛屿面积是多少?</p>
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<p><strong>岛屿</strong> 由一组上、下、左、右四个方向相连的<code>1</code> 形成。</p>
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<p><strong>岛屿</strong> 由一组上、下、左、右四个方向相连的&nbsp;<code>1</code> 形成。</p>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><strongclass="example">示例 1:</strong></p>
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<pre>
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<strong>输入: </strong>grid = [[1, 0], [0, 1]]
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<strong>输出:</strong> 3
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<strong>解释:</strong> 将一格0变成1,最终连通两个小岛得到面积为 3 的岛屿。
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</pre>
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<p><strong>示例 2:</strong></p>
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<p><strongclass="example">示例 2:</strong></p>
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<pre>
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<strong>输入: </strong>grid =<strong> </strong>[[1, 1], [1, 0]]
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<strong>输出:</strong> 4
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<strong>解释:</strong> 将一格0变成1,岛屿的面积扩大为 4。</pre>
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<p><strong>示例 3:</strong></p>
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<p><strongclass="example">示例 3:</strong></p>
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<pre>
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<strong>输入: </strong>grid = [[1, 1], [1, 1]]
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<strong>输出:</strong> 4
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<strong>解释:</strong> 没有0可以让我们变成1,面积依然为 4。</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>n == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 <= n <= 500</code></li>
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<li><code>1 &lt;= n &lt;= 500</code></li>
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<li><code>grid[i][j]</code> 为 <code>0</code> 或 <code>1</code></li>
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</ul>
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‎solution/1600-1699/1657.Determine if Two Strings Are Close/README.md‎

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<li>操作 1:交换任意两个 <strong>现有</strong> 字符。
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<ul>
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<li>例如,<code>a<strong>b</strong>cd<strong>e</strong> -> a<strong>e</strong>cd<strong>b</strong></code></li>
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<li>例如,<code>a<u>b</u>cd<u>e</u> -&gt; a<u>e</u>cd<u>b</u></code></li>
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</ul>
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</li>
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<li>操作 2:将一个 <strong>现有</strong> 字符的每次出现转换为另一个 <strong>现有</strong> 字符,并对另一个字符执行相同的操作。
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<ul>
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<li>例如,<code><strong>aa</strong>c<strong>abb</strong> -> <strong>bb</strong>c<strong>baa</strong></code>(所有 <code>a</code> 转化为 <code>b</code> ,而所有的 <code>b</code> 转换为 <code>a</code> )</li>
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<li>例如,<code><u>aa</u>c<u>abb</u> -&gt; <u>bb</u>c<u>baa</u></code>(所有 <code>a</code> 转化为 <code>b</code> ,而所有的 <code>b</code> 转换为 <code>a</code> )</li>
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</ul>
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<p>给你两个字符串,<code>word1</code> 和 <code>word2</code> 。如果<em> </em><code>word1</code><em> </em>和<em> </em><code>word2</code><em> </em><strong>接近 </strong>,就返回 <code>true</code> ;否则,返回<em> </em><code>false</code><em> </em>。</p>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>word1 = "abc", word2 = "bca"
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<strong>输出:</strong>true
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<strong>解释:</strong>2 次操作从 word1 获得 word2 。
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执行操作 1:"a<strong>bc</strong>" -> "a<strong>cb</strong>"
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执行操作 1:"<strong>a</strong>c<strong>b</strong>" -> "<strong>b</strong>c<strong>a</strong>"
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执行操作 1:"a<u>bc</u>" -&gt; "a<u>cb</u>"
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执行操作 1:"<u>a</u>c<u>b</u>" -&gt; "<u>b</u>c<u>a</u>"
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</pre>
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<p><strong>示例 2:</strong></p>
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<strong>输入:</strong>word1 = "cabbba", word2 = "abbccc"
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<strong>输出:</strong>true
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<strong>解释:</strong>3 次操作从 word1 获得 word2 。
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执行操作 1:"ca<strong>b</strong>bb<strong>a</strong>" -> "ca<strong>a</strong>bb<strong>b</strong>"
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执行操作 2:<code>"</code><strong>c</strong>aa<strong>bbb</strong>" -> "<strong>b</strong>aa<strong>ccc</strong>"
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执行操作 2:"<strong>baa</strong>ccc" -> "<strong>abb</strong>ccc"
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执行操作 1:"ca<u>b</u>bb<u>a</u>" -&gt; "ca<u>a</u>bb<u>b</u>"
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执行操作 2:<code>"</code><u>c</u>aa<u>bbb</u>" -&gt; "<u>b</u>aa<u>ccc</u>"
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执行操作 2:"<u>baa</u>ccc" -&gt; "<u>abb</u>ccc"
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre>
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<strong>输入:</strong>word1 = "cabbba", word2 = "aabbss"
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<strong>输出:</strong>false
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<strong>解释:</strong>不管执行多少次操作,都无法从 word1 得到 word2 ,反之亦然。</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<li><code>1 <= word1.length, word2.length <= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= word1.length, word2.length &lt;= 10<sup>5</sup></code></li>
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<li><code>word1</code> 和 <code>word2</code> 仅包含小写英文字母</li>
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</ul>
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‎solution/1600-1699/1657.Determine if Two Strings Are Close/README_EN.md‎

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<strong>Output:</strong> true
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<strong>Explanation:</strong> You can attain word2 from word1 in 3 operations.
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Apply Operation 1: &quot;ca<u>b</u>bb<u>a</u>&quot; -&gt; &quot;ca<u>a</u>bb<u>b</u>&quot;
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<code>Apply Operation 2: &quot;</code><u>c</u>aa<u>bbb</u>&quot; -&gt; &quot;<u>b</u>aa<u>ccc</u>&quot;
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Apply Operation 2: &quot;<u>c</u>aa<u>bbb</u>&quot; -&gt; &quot;<u>b</u>aa<u>ccc</u>&quot;
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Apply Operation 2: &quot;<u>baa</u>ccc&quot; -&gt; &quot;<u>abb</u>ccc&quot;
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</pre>
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‎solution/2100-2199/2186.Minimum Number of Steps to Make Two Strings Anagram II/README.md‎

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# [2186. 使两字符串互为字母异位词的最少步骤数](https://leetcode.cn/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii)
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# [2186. 制造字母异位词的最小步骤数 II](https://leetcode.cn/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii)
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[English Version](/solution/2100-2199/2186.Minimum%20Number%20of%20Steps%20to%20Make%20Two%20Strings%20Anagram%20II/README_EN.md)
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‎solution/2200-2299/2254.Design Video Sharing Platform/README_EN.md‎

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videoSharingPlatform.upload(&quot;789&quot;); // Since the video associated with videoId 0 was deleted,
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// 0 is the smallest available <code>videoId</code>, so return 0.
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videoSharingPlatform.watch(1, 0, 5); // The video associated with videoId 1 is &quot;456&quot;.
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// The video from minute 0 to min(5, 3 - 1) = 2 is &quot;456&quot;, so return &quot;453&quot;.
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// The video from minute 0 to min(5, 3 - 1) = 2 is &quot;456&quot;, so return &quot;456&quot;.
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videoSharingPlatform.watch(1, 0, 1); // The video associated with videoId 1 is &quot;456&quot;.
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// The video from minute 0 to min(1, 3 - 1) = 1 is &quot;45&quot;, so return &quot;45&quot;.
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videoSharingPlatform.like(1); // Increase the number of likes on the video associated with videoId 1.

‎solution/2600-2699/2654.Minimum Number of Operations to Make All Array Elements Equal to 1/README_EN.md‎

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<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
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</ul>
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<p>&nbsp;</p>
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<p><b>Follow-up:</b></p>
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<p>The <code>O(n)</code> time complexity&nbsp;solution works, but could you find an <code>O(1)</code> constant time complexity solution?</p>
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## Solutions
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### Solution 1

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