feat: update lc problems (doocs#2710)

Author: yanglbme

solution/0000-0099/0008.String to Integer (atoi)/README_EN.md

@@ -6,74 +6,108 @@
 
 ## Description
 
-<p>Implement the <code>myAtoi(string s)</code> function, which converts a string to a 32-bit signed integer (similar to C/C++&#39;s <code>atoi</code> function).</p>
+<p>Implement the <code>myAtoi(string s)</code> function, which converts a string to a 32-bit signed integer.</p>
 
 <p>The algorithm for <code>myAtoi(string s)</code> is as follows:</p>
 
 <ol>
-	<li>Read in and ignore any leading whitespace.</li>
-	<li>Check if the next character (if not already at the end of the string) is <code>&#39;-&#39;</code> or <code>&#39;+&#39;</code>. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.</li>
-	<li>Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.</li>
-	<li>Convert these digits into an integer (i.e. <code>&quot;123&quot; -&gt; 123</code>, <code>&quot;0032&quot; -&gt; 32</code>). If no digits were read, then the integer is <code>0</code>. Change the sign as necessary (from step 2).</li>
-	<li>If the integer is out of the 32-bit signed integer range <code>[-2<sup>31</sup>, 2<sup>31</sup> - 1]</code>, then clamp the integer so that it remains in the range. Specifically, integers less than <code>-2<sup>31</sup></code> should be clamped to <code>-2<sup>31</sup></code>, and integers greater than <code>2<sup>31</sup> - 1</code> should be clamped to <code>2<sup>31</sup> - 1</code>.</li>
-	<li>Return the integer as the final result.</li>
+	<li><strong>Whitespace</strong>: Ignore any leading whitespace (<code>&quot; &quot;</code>).</li>
+	<li><strong>Signedness</strong>: Determine the sign by checking if the next character is <code>&#39;-&#39;</code> or <code>&#39;+&#39;</code>, assuming positivity is neither present.</li>
+	<li><strong>Conversion</strong>: Read the integer by skipping leading zeros&nbsp;until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0.</li>
+	<li><strong>Edge case</strong>: If the integer is out of the 32-bit signed integer range <code>[-2<sup>31</sup>, 2<sup>31</sup> - 1]</code>, then round the integer to remain in the range. Specifically, integers less than <code>-2<sup>31</sup></code> should be rounded to <code>-2<sup>31</sup></code>, and integers greater than <code>2<sup>31</sup> - 1</code> should be rounded to <code>2<sup>31</sup> - 1</code>.</li>
 </ol>
 
-<p><strong>Note:</strong></p>
-
-<ul>
-	<li>Only the space character <code>&#39; &#39;</code> is considered a whitespace character.</li>
-	<li><strong>Do not ignore</strong> any characters other than the leading whitespace or the rest of the string after the digits.</li>
-</ul>
+<p>Return the integer as the final result.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;42&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">42</span></p>
+
+<p><strong>Explanation:</strong></p>
+
 <pre>
-<strong>Input:</strong> s = &quot;42&quot;
-<strong>Output:</strong> 42
-<strong>Explanation:</strong> The underlined characters are what is read in, the caret is the current reader position.
+The underlined characters are what is read in and the caret is the current reader position.
 Step 1: &quot;42&quot; (no characters read because there is no leading whitespace)
          ^
 Step 2: &quot;42&quot; (no characters read because there is neither a &#39;-&#39; nor &#39;+&#39;)
          ^
 Step 3: &quot;<u>42</u>&quot; (&quot;42&quot; is read in)
            ^
-The parsed integer is 42.
-Since 42 is in the range [-2<sup>31</sup>, 2<sup>31</sup> - 1], the final result is 42.
 </pre>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot; -042&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">-42</span></p>
+
+<p><strong>Explanation:</strong></p>
+
 <pre>
-<strong>Input:</strong> s = &quot;   -42&quot;
-<strong>Output:</strong> -42
-<strong>Explanation:</strong>
-Step 1: &quot;<u>   </u>-42&quot; (leading whitespace is read and ignored)
+Step 1: &quot;<u>   </u>-042&quot; (leading whitespace is read and ignored)
             ^
-Step 2: &quot;   <u>-</u>42&quot; (&#39;-&#39; is read, so the result should be negative)
+Step 2: &quot;   <u>-</u>042&quot; (&#39;-&#39; is read, so the result should be negative)
              ^
-Step 3: &quot;   -<u>42</u>&quot; (&quot;42&quot; is read in)
+Step 3: &quot;   -<u>042</u>&quot; (&quot;042&quot; is read in, leading zeros ignored in the result)
                ^
-The parsed integer is -42.
-Since -42 is in the range [-2<sup>31</sup>, 2<sup>31</sup> - 1], the final result is -42.
 </pre>
+</div>
 
 <p><strong class="example">Example 3:</strong></p>
 
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;1337c0d3&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1337</span></p>
+
+<p><strong>Explanation:</strong></p>
+
 <pre>
-<strong>Input:</strong> s = &quot;4193 with words&quot;
-<strong>Output:</strong> 4193
-<strong>Explanation:</strong>
-Step 1: &quot;4193 with words&quot; (no characters read because there is no leading whitespace)
+Step 1: &quot;1337c0d3&quot; (no characters read because there is no leading whitespace)
          ^
-Step 2: &quot;4193 with words&quot; (no characters read because there is neither a &#39;-&#39; nor &#39;+&#39;)
+Step 2: &quot;1337c0d3&quot; (no characters read because there is neither a &#39;-&#39; nor &#39;+&#39;)
          ^
-Step 3: &quot;<u>4193</u> with words&quot; (&quot;4193&quot; is read in; reading stops because the next character is a non-digit)
+Step 3: &quot;<u>1337</u>c0d3&quot; (&quot;1337&quot; is read in; reading stops because the next character is a non-digit)
              ^
-The parsed integer is 4193.
-Since 4193 is in the range [-2<sup>31</sup>, 2<sup>31</sup> - 1], the final result is 4193.
 </pre>
+</div>
+
+<p><strong class="example">Example 4:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;0-1&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<pre>
+Step 1: &quot;0-1&quot; (no characters read because there is no leading whitespace)
+         ^
+Step 2: &quot;0-1&quot; (no characters read because there is neither a &#39;-&#39; nor &#39;+&#39;)
+         ^
+Step 3: &quot;<u>0</u>-1&quot; (&quot;0&quot; is read in; reading stops because the next character is a non-digit)
+          ^
+</pre>
+</div>
+
+<p><strong class="example">Example 5:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;words and 987&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Reading stops at the first non-digit character &#39;w&#39;.</p>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/0000-0099/0012.Integer to Roman/README_EN.md

@@ -6,54 +6,107 @@
 
 ## Description
 
-<p>Roman numerals are represented by seven different symbols:&nbsp;<code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
-
-<pre>
-<strong>Symbol</strong>       <strong>Value</strong>
-I             1
-V             5
-X             10
-L             50
-C             100
-D             500
-M             1000</pre>
-
-<p>For example,&nbsp;<code>2</code> is written as <code>II</code>&nbsp;in Roman numeral, just two one&#39;s added together. <code>12</code> is written as&nbsp;<code>XII</code>, which is simply <code>X + II</code>. The number <code>27</code> is written as <code>XXVII</code>, which is <code>XX + V + II</code>.</p>
-
-<p>Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not <code>IIII</code>. Instead, the number four is written as <code>IV</code>. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as <code>IX</code>. There are six instances where subtraction is used:</p>
+<p>Seven different symbols represent Roman numerals with the following values:</p>
+
+<table>
+	<thead>
+		<tr>
+			<th>Symbol</th>
+			<th>Value</th>
+		</tr>
+	</thead>
+	<tbody>
+		<tr>
+			<td>I</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>V</td>
+			<td>5</td>
+		</tr>
+		<tr>
+			<td>X</td>
+			<td>10</td>
+		</tr>
+		<tr>
+			<td>L</td>
+			<td>50</td>
+		</tr>
+		<tr>
+			<td>C</td>
+			<td>100</td>
+		</tr>
+		<tr>
+			<td>D</td>
+			<td>500</td>
+		</tr>
+		<tr>
+			<td>M</td>
+			<td>1000</td>
+		</tr>
+	</tbody>
+</table>
+
+<p>Roman numerals are formed by appending&nbsp;the conversions of&nbsp;decimal place values&nbsp;from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:</p>
 
 <ul>
-	<li><code>I</code> can be placed before <code>V</code> (5) and <code>X</code> (10) to make 4 and 9.&nbsp;</li>
-	<li><code>X</code> can be placed before <code>L</code> (50) and <code>C</code> (100) to make 40 and 90.&nbsp;</li>
-	<li><code>C</code> can be placed before <code>D</code> (500) and <code>M</code> (1000) to make 400 and 900.</li>
+	<li>If the value does not start with 4 or&nbsp;9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.</li>
+	<li>If the value starts with 4 or 9 use the&nbsp;<strong>subtractive form</strong>&nbsp;representing&nbsp;one symbol subtracted from the following symbol, for example,&nbsp;4 is 1 (<code>I</code>) less than 5 (<code>V</code>): <code>IV</code>&nbsp;and 9 is 1 (<code>I</code>) less than 10 (<code>X</code>): <code>IX</code>.&nbsp;Only the following subtractive forms are used: 4 (<code>IV</code>), 9 (<code>IX</code>),&nbsp;40 (<code>XL</code>), 90 (<code>XC</code>), 400 (<code>CD</code>) and 900 (<code>CM</code>).</li>
+	<li>Only powers of 10 (<code>I</code>, <code>X</code>, <code>C</code>, <code>M</code>) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5&nbsp;(<code>V</code>), 50 (<code>L</code>), or 500 (<code>D</code>) multiple times. If you need to append a symbol&nbsp;4 times&nbsp;use the <strong>subtractive form</strong>.</li>
 </ul>
 
-<p>Given an integer, convert it to a roman numeral.</p>
+<p>Given an integer, convert it to a Roman numeral.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">num = 3749</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;MMMDCCXLIX&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
 <pre>
-<strong>Input:</strong> num = 3
-<strong>Output:</strong> &quot;III&quot;
-<strong>Explanation:</strong> 3 is represented as 3 ones.
+3000 = MMM as 1000 (M) + 1000 (M) + 1000 (M)
+ 700 = DCC as 500 (D) + 100 (C) + 100 (C)
+  40 = XL as 10 (X) less of 50 (L)
+   9 = IX as 1 (I) less of 10 (X)
+Note: 49 is not 1 (I) less of 50 (L) because the conversion is based on decimal places
 </pre>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">num = 58</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;LVIII&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
 <pre>
-<strong>Input:</strong> num = 58
-<strong>Output:</strong> &quot;LVIII&quot;
-<strong>Explanation:</strong> L = 50, V = 5, III = 3.
+50 = L
+ 8 = VIII
 </pre>
+</div>
 
 <p><strong class="example">Example 3:</strong></p>
 
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">num = 1994</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;MCMXCIV&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
 <pre>
-<strong>Input:</strong> num = 1994
-<strong>Output:</strong> &quot;MCMXCIV&quot;
-<strong>Explanation:</strong> M = 1000, CM = 900, XC = 90 and IV = 4.
+1000 = M
+ 900 = CM
+  90 = XC
+   4 = IV
 </pre>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/0000-0099/0038.Count and Say/README_EN.md

@@ -10,35 +10,42 @@
 
 <ul>
 	<li><code>countAndSay(1) = &quot;1&quot;</code></li>
-	<li><code>countAndSay(n)</code> is the way you would &quot;say&quot; the digit string from <code>countAndSay(n-1)</code>, which is then converted into a different digit string.</li>
+	<li><code>countAndSay(n)</code> is the run-length encoding of <code>countAndSay(n - 1)</code>.</li>
 </ul>
 
-<p>To determine how you &quot;say&quot; a digit string, split it into the <strong>minimal</strong> number of substrings such that each substring contains exactly <strong>one</strong> unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.</p>
+<p><a href="http://en.wikipedia.org/wiki/Run-length_encoding" target="_blank">Run-length encoding</a> (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string <code>&quot;3322251&quot;</code> we replace <code>&quot;33&quot;</code> with <code>&quot;23&quot;</code>, replace <code>&quot;222&quot;</code> with <code>&quot;32&quot;</code>, replace <code>&quot;5&quot;</code> with <code>&quot;15&quot;</code> and replace <code>&quot;1&quot;</code> with <code>&quot;11&quot;</code>. Thus the compressed string becomes <code>&quot;23321511&quot;</code>.</p>
 
-<p>For example, the saying and conversion for digit string <code>&quot;3322251&quot;</code>:</p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0038.Count%20and%20Say/images/countandsay.jpg" style="width: 581px; height: 172px;" />
-<p>Given a positive integer <code>n</code>, return <em>the </em><code>n<sup>th</sup></code><em> term of the <strong>count-and-say</strong> sequence</em>.</p>
+<p>Given a positive integer <code>n</code>, return <em>the </em><code>n<sup>th</sup></code><em> element of the <strong>count-and-say</strong> sequence</em>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
-<pre>
-<strong>Input:</strong> n = 1
-<strong>Output:</strong> &quot;1&quot;
-<strong>Explanation:</strong> This is the base case.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 4</span></p>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Output:</strong> <span class="example-io">&quot;1211&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
 
 <pre>
-<strong>Input:</strong> n = 4
-<strong>Output:</strong> &quot;1211&quot;
-<strong>Explanation:</strong>
 countAndSay(1) = &quot;1&quot;
-countAndSay(2) = say &quot;1&quot; = one 1 = &quot;11&quot;
-countAndSay(3) = say &quot;11&quot; = two 1&#39;s = &quot;21&quot;
-countAndSay(4) = say &quot;21&quot; = one 2 + one 1 = &quot;12&quot; + &quot;11&quot; = &quot;1211&quot;
+countAndSay(2) = RLE of &quot;1&quot; = &quot;11&quot;
+countAndSay(3) = RLE of &quot;11&quot; = &quot;21&quot;
+countAndSay(4) = RLE of &quot;21&quot; = &quot;1211&quot;
 </pre>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">n = 1</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;1&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>This is the base case.</p>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
@@ -47,6 +54,9 @@ countAndSay(4) = say &quot;21&quot; = one 2 + one 1 = &quot;12&quot; + &quot;11&
 	<li><code>1 &lt;= n &lt;= 30</code></li>
 </ul>
 
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Could you solve it iteratively?
+
 ## Solutions
 
 ### Solution 1: Simulation