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feat: update lc problems #3419

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2 changes: 1 addition & 1 deletion solution/0000-0099/0001.Two Sum/README.md
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<p>给定一个整数数组 <code>nums</code>&nbsp;和一个整数目标值 <code>target</code>，请你在该数组中找出 <strong>和为目标值 </strong><em><code>target</code></em>&nbsp; 的那&nbsp;<strong>两个</strong>&nbsp;整数，并返回它们的数组下标。</p>

<p>你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。</p>
<p>你可以假设每种输入只会对应一个答案，并且你不能使用两次相同的元素。</p>

<p>你可以按任意顺序返回答案。</p>

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4 changes: 2 additions & 2 deletions solution/0000-0099/0016.3Sum Closest/README.md
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<pre>
<strong>输入：</strong>nums = [-1,2,1,-4], target = 1
<strong>输出：</strong>2
<strong>解释：</strong>与 target 最接近的和是 2 (-1 + 2 + 1 = 2) 。
<strong>解释：</strong>与 target 最接近的和是 2 (-1 + 2 + 1 = 2)。
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>nums = [0,0,0], target = 1
<strong>输出：</strong>0
</pre>
<strong>解释：</strong>与 target 最接近的和是 0（0 + 0 + 0 = 0）。</pre>

<p>&nbsp;</p>

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84 changes: 57 additions & 27 deletions solution/0000-0099/0071.Simplify Path/README.md
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<!-- description:start -->

<p>给你一个字符串 <code>path</code> ，表示指向某一文件或目录的 Unix 风格 <strong>绝对路径 </strong>（以 <code>'/'</code> 开头），请你将其转化为更加简洁的规范路径。</p>
<p>给你一个字符串 <code>path</code> ，表示指向某一文件或目录的&nbsp;Unix 风格 <strong>绝对路径 </strong>（以 <code>'/'</code> 开头），请你将其转化为更加简洁的规范路径。</p>

<p class="MachineTrans-lang-zh-CN">在 Unix 风格的文件系统中，一个点（<code>.</code>）表示当前目录本身；此外，两个点 （<code>..</code>） 表示将目录切换到上一级（指向父目录）；两者都可以是复杂相对路径的组成部分。任意多个连续的斜杠（即，<code>'//'</code>）都被视为单个斜杠 <code>'/'</code> 。 对于此问题，任何其他格式的点（例如，<code>'...'</code>）均被视为文件/目录名称。</p>
<p class="MachineTrans-lang-zh-CN">在 Unix 风格的文件系统中，一个点（<code>.</code>）表示当前目录本身；此外，两个点 （<code>..</code>）&nbsp;表示将目录切换到上一级（指向父目录）；两者都可以是复杂相对路径的组成部分。任意多个连续的斜杠（即，<code>'//'</code>）都被视为单个斜杠 <code>'/'</code> 。 对于此问题，任何其他格式的点（例如，<code>'...'</code>）均被视为文件/目录名称。</p>

<p>请注意，返回的 <strong>规范路径</strong> 必须遵循下述格式：</p>

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<p>返回简化后得到的 <strong>规范路径</strong> 。</p>

<p> </p>
<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>
<p><strong class="example">示例 1：</strong></p>

<pre>
<strong>输入：</strong>path = "/home/"
<strong>输出：</strong>"/home"
<strong>解释：</strong>注意，最后一个目录名后面没有斜杠。 </pre>
<div class="example-block">
<p><span class="example-io"><b>输入：</b>path = "/home/"</span></p>

<p><strong>示例 2：</strong></p>
<p><span class="example-io"><b>输出：</b>"/home"</span></p>

<pre>
<strong>输入：</strong>path = "/../"
<strong>输出：</strong>"/"
<strong>解释：</strong>从根目录向上一级是不可行的，因为根目录是你可以到达的最高级。
</pre>
<p><strong>解释：</strong></p>

<p><strong>示例 3：</strong></p>
<p>应删除尾部斜杠。</p>
</div>

<pre>
<strong>输入：</strong>path = "/home//foo/"
<strong>输出：</strong>"/home/foo"
<strong>解释：</strong>在规范路径中，多个连续斜杠需要用一个斜杠替换。
</pre>
<p><strong class="example">示例 2：</strong></p>

<p><strong>示例 4：</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入：</b></span><span class="example-io">path = "/home//foo/"</span></p>

<pre>
<strong>输入：</strong>path = "/a/./b/../../c/"
<strong>输出：</strong>"/c"
</pre>
<p><span class="example-io"><b>输出：</b></span><span class="example-io">"/home/foo"</span></p>

<p> </p>
<p><strong>解释：</strong></p>

<p>多个连续的斜杠被单个斜杠替换。</p>
</div>

<p><strong class="example">示例 3：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b></span><span class="example-io">path = "/home/user/Documents/../Pictures"</span></p>

<p><span class="example-io"><b>输出：</b></span><span class="example-io">"/home/user/Pictures"</span></p>

<p><strong>解释：</strong></p>

<p>两个点&nbsp;<code>".."</code>&nbsp;表示上一级目录。</p>
</div>

<p><strong class="example">示例 4：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b></span><span class="example-io">path = "/../"</span></p>

<p><span class="example-io"><b>输出：</b></span><span class="example-io">"/"</span></p>

<p><strong>解释：</strong></p>

<p>不可能从根目录上升级一级。</p>
</div>

<p><strong class="example">示例 5：</strong></p>

<div class="example-block">
<p><span class="example-io"><b>输入：</b></span><span class="example-io">path = "/.../a/../b/c/../d/./"</span></p>

<p><span class="example-io"><b>输出：</b></span><span class="example-io">"/.../b/d"</span></p>

<p><strong>解释：</strong></p>

<p><code>"..."</code> 是此问题中目录的有效名称。</p>
</div>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
<li><code>1 <= path.length <= 3000</code></li>
<li><code>1 &lt;= path.length &lt;= 3000</code></li>
<li><code>path</code> 由英文字母，数字，<code>'.'</code>，<code>'/'</code> 或 <code>'_'</code> 组成。</li>
<li><code>path</code> 是一个有效的 Unix 风格绝对路径。</li>
</ul>
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2 changes: 1 addition & 1 deletion solution/0200-0299/0230.Kth Smallest Element in a BST/README.md
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<!-- problem:start -->

# [230. 二叉搜索树中第K小的元素](https://leetcode.cn/problems/kth-smallest-element-in-a-bst)
# [230. 二叉搜索树中第 K 小的元素](https://leetcode.cn/problems/kth-smallest-element-in-a-bst)

[English Version](/solution/0200-0299/0230.Kth%20Smallest%20Element%20in%20a%20BST/README_EN.md)

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2 changes: 1 addition & 1 deletion solution/0300-0399/0308.Range Sum Query 2D - Mutable/README.md
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<!-- problem:start -->

# [308. 二维区域和检索 - 可变 🔒](https://leetcode.cn/problems/range-sum-query-2d-mutable)
# [308. 二维区域和检索 - 矩阵可修改 🔒](https://leetcode.cn/problems/range-sum-query-2d-mutable)

[English Version](/solution/0300-0399/0308.Range%20Sum%20Query%202D%20-%20Mutable/README_EN.md)

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30 changes: 17 additions & 13 deletions solution/0400-0499/0457.Circular Array Loop/README.md
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<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<p><strong class="example">示例 1：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0457.Circular%20Array%20Loop/images/1723688159-qYjpWT-image.png" style="width: 402px; height: 289px;" />
<pre>
<strong>输入：</strong>nums = [2,-1,1,2,2]
<strong>输出：</strong>true
<strong>解释：</strong>存在循环，按下标 0 -&gt; 2 -&gt; 3 -&gt; 0 。循环长度为 3 。
<strong>解释：</strong>图片展示了节点间如何连接。白色节点向前跳跃，而红色节点向后跳跃。
我们可以看到存在循环，按下标 0 -&gt; 2 -&gt; 3 -&gt; 0 --&gt; ...，并且其中的所有节点都是白色（以相同方向跳跃）。
</pre>

<p><strong>示例 2：</strong></p>

<p><strong class="example">示例 2：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0457.Circular%20Array%20Loop/images/1723688183-lRSkjp-image.png" style="width: 402px; height: 390px;" />
<pre>
<strong>输入：</strong>nums = [-1,2]
<strong>输入：</strong>nums = [-1,-2,-3,-4,-5,6]
<strong>输出：</strong>false
<strong>解释：</strong>按下标 1 -&gt; 1 -&gt; 1 ... 的运动无法构成循环，因为循环的长度为 1 。根据定义，循环的长度必须大于 1 。
<strong>解释：</strong>图片展示了节点间如何连接。白色节点向前跳跃，而红色节点向后跳跃。
唯一的循环长度为 1，所以返回 false。
</pre>

<p><strong>示例 3:</strong></p>

<p><strong class="example">示例 3：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0457.Circular%20Array%20Loop/images/1723688199-nhaMuF-image.png" style="width: 497px; height: 242px;" />
<pre>
<strong>输入：</strong>nums = [-2,1,-1,-2,-2]
<strong>输出：</strong>false
<strong>解释：</strong>按下标 1 -&gt; 2 -&gt; 1 -&gt; ... 的运动无法构成循环，因为 nums[1] 是正数，而 nums[2] 是负数。
所有 nums[seq[j]] 应当不是全正就是全负。</pre>
<strong>输入：</strong>nums = [1,-1,5,1,4]
<strong>输出：</strong>true
<strong>解释：</strong>图片展示了节点间如何连接。白色节点向前跳跃，而红色节点向后跳跃。
我们可以看到存在循环，按下标 0 --&gt; 1 --&gt; 0 --&gt; ...，当它的大小大于 1 时，它有一个向前跳的节点和一个向后跳的节点，所以 <strong>它不是一个循环</strong>。
我们可以看到存在循环，按下标 3 --&gt; 4 --&gt; 3 --&gt; ...，并且其中的所有节点都是白色（以相同方向跳跃）。
</pre>

<p>&nbsp;</p>

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17 changes: 8 additions & 9 deletions solution/0600-0699/0638.Shopping Offers/README.md
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<p>返回<strong> 确切 </strong>满足购物清单所需花费的最低价格，你可以充分利用大礼包的优惠活动。你不能购买超出购物清单指定数量的物品，即使那样会降低整体价格。任意大礼包可无限次购买。</p>

<p> </p>
<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

Expand All @@ -51,19 +51,18 @@ tags:
需要买 1A ，2B 和 1C ，所以付 ¥4 买 1A 和 1B（大礼包 1），以及 ¥3 购买 1B ， ¥4 购买 1C 。
不可以购买超出待购清单的物品，尽管购买大礼包 2 更加便宜。</pre>

<p> </p>
<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
<li><code>n == price.length</code></li>
<li><code>n == needs.length</code></li>
<li><code>1 <= n <= 6</code></li>
<li><code>0 <= price[i] <= 10</code></li>
<li><code>0 <= needs[i] <= 10</code></li>
<li><code>1 <= special.length <= 100</code></li>
<li><code>n == price.length == needs.length</code></li>
<li><code>1 &lt;= n &lt;= 6</code></li>
<li><code>0 &lt;= price[i], needs[i] &lt;= 10</code></li>
<li><code>1 &lt;= special.length &lt;= 100</code></li>
<li><code>special[i].length == n + 1</code></li>
<li><code>0 <= special[i][j] <= 50</code></li>
<li><code>0 &lt;= special[i][j] &lt;= 50</code></li>
<li>生成的输入对于&nbsp;<code>0 &lt;= j &lt;= n - 1</code> 至少有一个&nbsp;<code>special[i][j]</code>&nbsp;非零。</li>
</ul>

<!-- description:end -->
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1 change: 1 addition & 0 deletions solution/0600-0699/0638.Shopping Offers/README_EN.md
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<li><code>1 &lt;= special.length &lt;= 100</code></li>
<li><code>special[i].length == n + 1</code></li>
<li><code>0 &lt;= special[i][j] &lt;= 50</code></li>
<li>The input is generated that at least one of <code>special[i][j]</code> is non-zero for <code>0 &lt;= j &lt;= n - 1</code>.</li>
</ul>

<!-- description:end -->
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2 changes: 1 addition & 1 deletion solution/0700-0799/0777.Swap Adjacent in LR String/README_EN.md
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<!-- description:start -->

<p>In a string composed of <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;X&#39;</code> characters, like <code>&quot;RXXLRXRXL&quot;</code>, a move consists of either replacing one occurrence of <code>&quot;XL&quot;</code> with <code>&quot;LX&quot;</code>, or replacing one occurrence of <code>&quot;RX&quot;</code> with <code>&quot;XR&quot;</code>. Given the starting string <code>start</code> and the ending string <code>end</code>, return <code>True</code> if and only if there exists a sequence of moves to transform one string to the other.</p>
<p>In a string composed of <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;X&#39;</code> characters, like <code>&quot;RXXLRXRXL&quot;</code>, a move consists of either replacing one occurrence of <code>&quot;XL&quot;</code> with <code>&quot;LX&quot;</code>, or replacing one occurrence of <code>&quot;RX&quot;</code> with <code>&quot;XR&quot;</code>. Given the starting string <code>start</code> and the ending string <code>end</code>, return <code>True</code> if and only if there exists a sequence of moves to transform <code>start</code> to <code>end</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
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40 changes: 22 additions & 18 deletions solution/0700-0799/0787.Cheapest Flights Within K Stops/README.md
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<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0787.Cheapest%20Flights%20Within%20K%20Stops/images/cheapest-flights-within-k-stops-3drawio.png" style="width: 332px; height: 392px;" />
<pre>
<strong>输入:</strong>
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
<strong>输出:</strong> 200
<strong>解释:</strong>
城市航班图如下
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0787.Cheapest%20Flights%20Within%20K%20Stops/images/995.png" style="height: 180px; width: 246px;" />

从城市 0 到城市 2 在 1 站中转以内的最便宜价格是 200，如图中红色所示。</pre>
n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
<strong>输出:</strong> 700
<strong>解释:</strong> 城市航班图如上
从城市 0 到城市 3 经过最多 1 站的最佳路径用红色标记，费用为 100 + 600 = 700。
请注意，通过城市 [0, 1, 2, 3] 的路径更便宜，但无效，因为它经过了 2 站。
</pre>

<p><strong>示例 2：</strong></p>

<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0787.Cheapest%20Flights%20Within%20K%20Stops/images/cheapest-flights-within-k-stops-1drawio.png" style="width: 332px; height: 242px;" />
<pre>
<strong>输入:</strong>
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
<strong>输出:</strong> 500
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
<strong>输出:</strong> 200
<strong>解释:</strong>
城市航班图如下
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0787.Cheapest%20Flights%20Within%20K%20Stops/images/995.png" style="height: 180px; width: 246px;" />

从城市 0 到城市 2 在 0 站中转以内的最便宜价格是 500，如图中蓝色所示。</pre>
城市航班图如上
从城市 0 到城市 2 经过最多 1 站的最佳路径标记为红色，费用为 100 + 100 = 200。
</pre>

<p>&nbsp;</p>
<p><strong class="example">示例 3：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0787.Cheapest%20Flights%20Within%20K%20Stops/images/cheapest-flights-within-k-stops-2drawio.png" style="width: 332px; height: 242px;" />
<pre>
<b>输入：</b>n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
<b>输出：</b>500
<strong>解释：</strong>
城市航班图如上
从城市 0 到城市 2 不经过站点的最佳路径标记为红色，费用为 500。
</pre>

<p><strong>提示：</strong></p>

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2 changes: 1 addition & 1 deletion solution/0800-0899/0841.Keys and Rooms/README.md
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<p>有 <code>n</code> 个房间，房间按从 <code>0</code> 到 <code>n - 1</code> 编号。最初，除 <code>0</code> 号房间外的其余所有房间都被锁住。你的目标是进入所有的房间。然而，你不能在没有获得钥匙的时候进入锁住的房间。</p>

<p>当你进入一个房间，你可能会在里面找到一套不同的钥匙，每把钥匙上都有对应的房间号，即表示钥匙可以打开的房间。你可以拿上所有钥匙去解锁其他房间。</p>
<p>当你进入一个房间，你可能会在里面找到一套 <strong>不同的钥匙</strong>，每把钥匙上都有对应的房间号，即表示钥匙可以打开的房间。你可以拿上所有钥匙去解锁其他房间。</p>

<p>给你一个数组 <code>rooms</code> 其中 <code>rooms[i]</code> 是你进入 <code>i</code> 号房间可以获得的钥匙集合。如果能进入 <strong>所有</strong> 房间返回 <code>true</code>，否则返回 <code>false</code>。</p>

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2 changes: 1 addition & 1 deletion solution/0900-0999/0919.Complete Binary Tree Inserter/README.md
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Expand Up @@ -21,7 +21,7 @@ tags:

<p><strong>完全二叉树</strong> 是每一层（除最后一层外）都是完全填充（即，节点数达到最大）的，并且所有的节点都尽可能地集中在左侧。</p>

<p>设计一种算法，将一个新节点插入到一个完整的二叉树中，并在插入后保持其完整。</p>
<p>设计一种算法，将一个新节点插入到一棵完全二叉树中，并在插入后保持其完整。</p>

<p>实现 <code>CBTInserter</code> 类:</p>

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15 changes: 6 additions & 9 deletions solution/1000-1099/1032.Stream of Characters/README.md
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Expand Up @@ -202,14 +202,11 @@ class StreamChecker {

```cpp
class Trie {
public:
vector<Trie*> children;
bool isEnd;

Trie()
: children(26)
, isEnd(false) {}
private:
Trie* children[26]{};
bool isEnd = false;

public:
void insert(string& w) {
Trie* node = this;
reverse(w.begin(), w.end());
Expand All @@ -225,7 +222,7 @@ public:

bool search(string& w) {
Trie* node = this;
for (int i = w.size() - 1; ~i; --i) {
for (int i = w.size() - 1, j = 0; ~i && j < 201; --i, ++j) {
int idx = w[i] - 'a';
if (!node->children[idx]) {
return false;
Expand All @@ -245,7 +242,7 @@ public:
string s;

StreamChecker(vector<string>& words) {
for (auto&& w : words) {
for (auto& w : words) {
trie->insert(w);
}
}
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