feat: update lc problems

Author: yanglbme

solution/0000-0099/0015.3Sum/README_EN.md

@@ -10,20 +10,39 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<pre><strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
+
+<pre>
+<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
 <strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
-</pre><p><strong>Example 2:</strong></p>
-<pre><strong>Input:</strong> nums = []
-<strong>Output:</strong> []
-</pre><p><strong>Example 3:</strong></p>
-<pre><strong>Input:</strong> nums = [0]
+<strong>Explanation:</strong> 
+nums[0] + nums[1] + nums[1] = (-1) + 0 + 1 = 0.
+nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
+nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
+The distinct triplets are [-1,0,1] and [-1,-1,2].
+Notice that the order of the output and the order of the triplets does not matter.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,1,1]
 <strong>Output:</strong> []
+<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
 </pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,0,0]
+<strong>Output:</strong> [[0,0,0]]
+<strong>Explanation:</strong> The only possible triplet sums up to 0.
+</pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>0 &lt;= nums.length &lt;= 3000</code></li>
+	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
 	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
 </ul>
 

solution/0100-0199/0155.Min Stack/README_EN.md

@@ -16,6 +16,8 @@
 	<li><code>int getMin()</code> retrieves the minimum element in the stack.</li>
 </ul>
 
+<p>You must implement a solution with <code>O(1)</code> time complexity for each function.</p>
+
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 

solution/0100-0199/0162.Find Peak Element/README_EN.md

@@ -6,11 +6,11 @@
 
 <p>A peak element is an element that is strictly greater than its neighbors.</p>
 
-<p>Given an integer array <code>nums</code>, find a peak element, and return its index. If&nbsp;the array contains multiple peaks, return the index to <strong>any of the peaks</strong>.</p>
+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, find a peak element, and return its index. If the array contains multiple peaks, return the index to <strong>any of the peaks</strong>.</p>
 
-<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>.</p>
+<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.</p>
 
-<p>You must write an algorithm that runs in&nbsp;<code>O(log n)</code> time.</p>
+<p>You must write an algorithm that runs in <code>O(log n)</code> time.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0200-0299/0223.Rectangle Area/README_EN.md

@@ -29,7 +29,10 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>-10<sup>4</sup> &lt;= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= ax1 &lt;= ax2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= ay1 &lt;= ay2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= bx1 &lt;= bx2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= by1 &lt;= by2 &lt;= 10<sup>4</sup></code></li>
 </ul>
 
 ## Solutions

solution/0300-0399/0320.Generalized Abbreviation/README.md

@@ -10,6 +10,7 @@
 
 <ul>
 	<li>例如，<code>"abcde"</code> 可以缩写为：
+
     <ul>
     	<li><code>"a3e"</code>（<code>"bcd"</code> 变为 <code>"3"</code> ）</li>
     	<li><code>"1bcd1"</code>（<code>"a"</code> 和 <code>"e"</code> 都变为 <code>"1"</code>）<meta charset="UTF-8" /></li>
@@ -23,6 +24,7 @@
     	<li><meta charset="UTF-8" /><code>"22de"</code>&nbsp;(<code>"ab"</code> 变为&nbsp;<code>"2"</code>&nbsp;，&nbsp;<code>"bc"</code>&nbsp;变为&nbsp;<code>"2"</code>) &nbsp;是无效的，因为被选择的字符串是重叠的</li>
     </ul>
     </li>
+
 </ul>
 
 <p>给你一个字符串&nbsp;<code>word</code> ，返回&nbsp;<em>一个由</em>&nbsp;<code>word</code> 的<em>所有可能 <strong>广义缩写词</strong> 组成的列表</em>&nbsp;。按 <strong>任意顺序</strong> 返回答案。</p>

solution/0300-0399/0336.Palindrome Pairs/README.md

@@ -32,6 +32,8 @@
 <strong>输出：</strong>[[0,1],[1,0]]
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0554.Brick Wall/README.md

@@ -28,6 +28,8 @@
 <strong>输出：</strong>3
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0560.Subarray Sum Equals K/README.md

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个整数数组 <code>nums</code> 和一个整数&nbsp;<code>k</code> ，请你统计并返回 <em>该数组中和为&nbsp;<code>k</code><strong>&nbsp;</strong>的子数组的个数&nbsp;</em>。</p>
+<p>给你一个整数数组 <code>nums</code> 和一个整数&nbsp;<code>k</code> ，请你统计并返回 <em>该数组中和为&nbsp;<code>k</code><strong>&nbsp;</strong>的连续子数组的个数&nbsp;</em>。</p>
 
 <p>&nbsp;</p>
 

solution/0500-0599/0561.Array Partition/README.md

@@ -41,7 +41,6 @@
 	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/0500-0599/0561.Array Partition/README_EN.md

@@ -35,7 +35,6 @@ So the maximum possible sum is 4.</pre>
 	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
-
 ## Solutions
 
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