feat: add new lc problems

solution/0100-0199/0131.Palindrome Partitioning/README.md

@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个字符串 <code>s</code>，请你将<em> </em><code>s</code><em> </em>分割成一些 <span data-keyword="substring-nonempty">子串</span>，使每个子串都是 <strong><span data-keyword="palindrome-string">回文串</span></strong> 。返回 <code>s</code> 所有可能的分割方案。</p>
+<p>给你一个字符串 <code>s</code>，请你将<em> </em><code>s</code><em> </em>分割成一些子串，使每个子串都是 <strong><span data-keyword="palindrome-string">回文串</span></strong> 。返回 <code>s</code> 所有可能的分割方案。</p>
 
 <p>&nbsp;</p>
 

solution/0100-0199/0195.Tenth Line/README_EN.md

@@ -23,26 +23,41 @@ tags:
 <p>Assume that <code>file.txt</code> has the following content:</p>
 
 <pre>
+
 Line 1
+
 Line 2
+
 Line 3
+
 Line 4
+
 Line 5
+
 Line 6
+
 Line 7
+
 Line 8
+
 Line 9
+
 Line 10
+
 </pre>
 
 <p>Your script should output the tenth line, which is:</p>
 
 <pre>
+
 Line 10
+
 </pre>
 
 <div class="spoilers"><b>Note:</b><br />
+
 1. If the file contains less than 10 lines, what should you output?<br />
+
 2. There&#39;s at least three different solutions. Try to explore all possibilities.</div>
 
 <!-- description:end -->

solution/0200-0299/0278.First Bad Version/README.md

@@ -30,8 +30,8 @@ tags:
 <strong>输入：</strong>n = 5, bad = 4
 <strong>输出：</strong>4
 <strong>解释：</strong>
-<code>调用 isBadVersion(3) -&gt; false
-调用 isBadVersion(5)&nbsp;-&gt; true
+<code>调用 isBadVersion(3) -&gt; false 
+调用 isBadVersion(5)&nbsp;-&gt; true 
 调用 isBadVersion(4)&nbsp;-&gt; true</code>
 <code>所以，4 是第一个错误的版本。</code>
 </pre>

solution/0500-0599/0525.Contiguous Array/README.md

@@ -20,35 +20,28 @@ tags:
 
 <p>给定一个二进制数组 <code>nums</code> , 找到含有相同数量的 <code>0</code> 和 <code>1</code> 的最长连续子数组，并返回该子数组的长度。</p>
 
-<p>&nbsp;</p>
+<p> </p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong>示例 1:</strong></p>
 
 <pre>
-<strong>输入：</strong>nums = [0,1]
-<strong>输出：</strong>2
-<strong>说明：</strong>[0, 1] 是具有相同数量 0 和 1 的最长连续子数组。</pre>
+<strong>输入:</strong> nums = [0,1]
+<strong>输出:</strong> 2
+<strong>说明:</strong> [0, 1] 是具有相同数量 0 和 1 的最长连续子数组。</pre>
 
-<p><strong>示例 2：</strong></p>
+<p><strong>示例 2:</strong></p>
 
 <pre>
-<strong>输入：</strong>nums = [0,1,0]
-<strong>输出：</strong>2
-<strong>说明：</strong>[0, 1] (或 [1, 0]) 是具有相同数量 0 和 1 的最长连续子数组。</pre>
+<strong>输入:</strong> nums = [0,1,0]
+<strong>输出:</strong> 2
+<strong>说明:</strong> [0, 1] (或 [1, 0]) 是具有相同数量0和1的最长连续子数组。</pre>
 
-<p><strong>示例 3：</strong></p>
-
-<pre>
-<b>输入：</b>nums = [0,1,1,1,1,1,0,0,0]
-<b>输出：</b>6
-<b>解释：</b>[1,1,1,0,0,0] 是具有相同数量 0 和 1 的最长连续子数组。</pre>
-
-<p>&nbsp;</p>
+<p> </p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
 	<li><code>nums[i]</code> 不是 <code>0</code> 就是 <code>1</code></li>
 </ul>
 

solution/0500-0599/0525.Contiguous Array/README_EN.md

@@ -37,14 +37,6 @@ tags:
 <strong>Explanation:</strong> [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
-
-<pre>
-<strong>Input:</strong> nums = [0,1,1,1,1,1,0,0,0]
-<strong>Output:</strong> 6
-<strong>Explanation:</strong> [1,1,1,0,0,0] is the longest contiguous subarray with equal number of 0 and 1.
-</pre>
-
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

solution/0700-0799/0766.Toeplitz Matrix/README.md

@@ -29,8 +29,8 @@ tags:
 <strong>输入：</strong>matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]]
 <strong>输出：</strong>true
 <strong>解释：</strong>
-在上述矩阵中, 其对角线为:
-"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]"。
+在上述矩阵中, 其对角线为: 
+"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]"。 
 各条对角线上的所有元素均相同, 因此答案是 True 。
 </pre>
 

solution/0700-0799/0799.Champagne Tower/README_EN.md

@@ -27,35 +27,51 @@ tags:
 <p>Now after pouring some non-negative integer cups of champagne, return how full the <code>j<sup>th</sup></code> glass in the <code>i<sup>th</sup></code> row is (both <code>i</code> and <code>j</code> are 0-indexed.)</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> poured = 1, query_row = 1, query_glass = 1
+
 <strong>Output:</strong> 0.00000
+
 <strong>Explanation:</strong> We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.
+
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> poured = 2, query_row = 1, query_glass = 1
+
 <strong>Output:</strong> 0.50000
+
 <strong>Explanation:</strong> We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.
+
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> poured = 100000009, query_row = 33, query_glass = 17
+
 <strong>Output:</strong> 1.00000
+
 </pre>
 
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
-	<li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
+
+    <li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
+
+    <li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
+
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1108.Defanging an IP Address/README_EN.md

@@ -23,18 +23,29 @@ tags:
 <p>A <em>defanged&nbsp;IP address</em>&nbsp;replaces every period <code>&quot;.&quot;</code> with <code>&quot;[.]&quot;</code>.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
+
 <pre><strong>Input:</strong> address = "1.1.1.1"
+
 <strong>Output:</strong> "1[.]1[.]1[.]1"
+
 </pre><p><strong class="example">Example 2:</strong></p>
+
 <pre><strong>Input:</strong> address = "255.100.50.0"
+
 <strong>Output:</strong> "255[.]100[.]50[.]0"
+
 </pre>
+
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li>The given <code>address</code> is a valid IPv4 address.</li>
+
+    <li>The given <code>address</code> is a valid IPv4 address.</li>
+
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1111.Maximum Nesting Depth of Two Valid Parentheses Strings/README_EN.md

@@ -22,17 +22,25 @@ tags:
 <p>A string is a <em>valid parentheses string</em>&nbsp;(denoted VPS) if and only if it consists of <code>&quot;(&quot;</code> and <code>&quot;)&quot;</code> characters only, and:</p>
 
 <ul>
-	<li>It is the empty string, or</li>
-	<li>It can be written as&nbsp;<code>AB</code>&nbsp;(<code>A</code>&nbsp;concatenated with&nbsp;<code>B</code>), where&nbsp;<code>A</code>&nbsp;and&nbsp;<code>B</code>&nbsp;are VPS&#39;s, or</li>
-	<li>It can be written as&nbsp;<code>(A)</code>, where&nbsp;<code>A</code>&nbsp;is a VPS.</li>
+
+    <li>It is the empty string, or</li>
+
+    <li>It can be written as&nbsp;<code>AB</code>&nbsp;(<code>A</code>&nbsp;concatenated with&nbsp;<code>B</code>), where&nbsp;<code>A</code>&nbsp;and&nbsp;<code>B</code>&nbsp;are VPS&#39;s, or</li>
+
+    <li>It can be written as&nbsp;<code>(A)</code>, where&nbsp;<code>A</code>&nbsp;is a VPS.</li>
+
 </ul>
 
 <p>We can&nbsp;similarly define the <em>nesting depth</em> <code>depth(S)</code> of any VPS <code>S</code> as follows:</p>
 
 <ul>
-	<li><code>depth(&quot;&quot;) = 0</code></li>
-	<li><code>depth(A + B) = max(depth(A), depth(B))</code>, where <code>A</code> and <code>B</code> are VPS&#39;s</li>
-	<li><code>depth(&quot;(&quot; + A + &quot;)&quot;) = 1 + depth(A)</code>, where <code>A</code> is a VPS.</li>
+
+    <li><code>depth(&quot;&quot;) = 0</code></li>
+
+    <li><code>depth(A + B) = max(depth(A), depth(B))</code>, where <code>A</code> and <code>B</code> are VPS&#39;s</li>
+
+    <li><code>depth(&quot;(&quot; + A + &quot;)&quot;) = 1 + depth(A)</code>, where <code>A</code> is a VPS.</li>
+
 </ul>
 
 <p>For example,&nbsp; <code>&quot;&quot;</code>,&nbsp;<code>&quot;()()&quot;</code>, and&nbsp;<code>&quot;()(()())&quot;</code>&nbsp;are VPS&#39;s (with nesting depths 0, 1, and 2), and <code>&quot;)(&quot;</code> and <code>&quot;(()&quot;</code> are not VPS&#39;s.</p>

solution/1100-1199/1138.Alphabet Board Path/README_EN.md

@@ -28,31 +28,49 @@ tags:
 <p>We may make the following moves:</p>
 
 <ul>
-	<li><code>&#39;U&#39;</code> moves our position up one row, if the position exists on the board;</li>
-	<li><code>&#39;D&#39;</code> moves our position down one row, if the position exists on the board;</li>
-	<li><code>&#39;L&#39;</code> moves our position left one column, if the position exists on the board;</li>
-	<li><code>&#39;R&#39;</code> moves our position right one column, if the position exists on the board;</li>
-	<li><code>&#39;!&#39;</code>&nbsp;adds the character <code>board[r][c]</code> at our current position <code>(r, c)</code>&nbsp;to the&nbsp;answer.</li>
+
+    <li><code>&#39;U&#39;</code> moves our position up one row, if the position exists on the board;</li>
+
+    <li><code>&#39;D&#39;</code> moves our position down one row, if the position exists on the board;</li>
+
+    <li><code>&#39;L&#39;</code> moves our position left one column, if the position exists on the board;</li>
+
+    <li><code>&#39;R&#39;</code> moves our position right one column, if the position exists on the board;</li>
+
+    <li><code>&#39;!&#39;</code>&nbsp;adds the character <code>board[r][c]</code> at our current position <code>(r, c)</code>&nbsp;to the&nbsp;answer.</li>
+
 </ul>
 
 <p>(Here, the only positions that exist on the board are positions with letters on them.)</p>
 
 <p>Return a sequence of moves that makes our answer equal to <code>target</code>&nbsp;in the minimum number of moves.&nbsp; You may return any path that does so.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
+
 <pre><strong>Input:</strong> target = "leet"
+
 <strong>Output:</strong> "DDR!UURRR!!DDD!"
+
 </pre><p><strong class="example">Example 2:</strong></p>
+
 <pre><strong>Input:</strong> target = "code"
+
 <strong>Output:</strong> "RR!DDRR!UUL!R!"
+
 </pre>
+
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= target.length &lt;= 100</code></li>
-	<li><code>target</code> consists only of English lowercase letters.</li>
+
+    <li><code>1 &lt;= target.length &lt;= 100</code></li>
+
+    <li><code>target</code> consists only of English lowercase letters.</li>
+
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1139.Largest 1-Bordered Square/README_EN.md

@@ -23,27 +23,39 @@ tags:
 <p>Given a 2D <code>grid</code> of <code>0</code>s and <code>1</code>s, return the number of elements in&nbsp;the largest <strong>square</strong>&nbsp;subgrid that has all <code>1</code>s on its <strong>border</strong>, or <code>0</code> if such a subgrid&nbsp;doesn&#39;t exist in the <code>grid</code>.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> grid = [[1,1,1],[1,0,1],[1,1,1]]
+
 <strong>Output:</strong> 9
+
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> grid = [[1,1,0,0]]
+
 <strong>Output:</strong> 1
+
 </pre>
 
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= grid.length &lt;= 100</code></li>
-	<li><code>1 &lt;= grid[0].length &lt;= 100</code></li>
-	<li><code>grid[i][j]</code> is <code>0</code> or <code>1</code></li>
+
+    <li><code>1 &lt;= grid.length &lt;= 100</code></li>
+
+    <li><code>1 &lt;= grid[0].length &lt;= 100</code></li>
+
+    <li><code>grid[i][j]</code> is <code>0</code> or <code>1</code></li>
+
 </ul>
 
 <!-- description:end -->