feat: update lc problems

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README.md

@@ -27,7 +27,7 @@ tags:
 <pre>
 <strong>输入: </strong>s = "abcabcbb"
 <strong>输出: </strong>3 
-<strong>解释:</strong> 因为无重复字符的最长子串是 <code>"abc"</code>，所以其长度为 3。
+<strong>解释:</strong> 因为无重复字符的最长子串是 <code>"abc"</code>，所以其长度为 3。注意 "bca" 和 "cab" 也是正确答案。
 </pre>
 
 <p><strong>示例 2:</strong></p>

solution/0000-0099/0017.Letter Combinations of a Phone Number/README.md

@@ -35,13 +35,6 @@ tags:
 
 <p><strong>示例 2：</strong></p>
 
-<pre>
-<strong>输入：</strong>digits = ""
-<strong>输出：</strong>[]
-</pre>
-
-<p><strong>示例 3：</strong></p>
-
 <pre>
 <strong>输入：</strong>digits = "2"
 <strong>输出：</strong>["a","b","c"]
@@ -52,7 +45,7 @@ tags:
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>0 &lt;= digits.length &lt;= 4</code></li>
+	<li><code>1 &lt;= digits.length &lt;= 4</code></li>
 	<li><code>digits[i]</code> 是范围 <code>['2', '9']</code> 的一个数字。</li>
 </ul>
 

solution/0000-0099/0017.Letter Combinations of a Phone Number/README_EN.md

@@ -32,13 +32,6 @@ tags:
 
 <p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> digits = &quot;&quot;
-<strong>Output:</strong> []
-</pre>
-
-<p><strong class="example">Example 3:</strong></p>
-
 <pre>
 <strong>Input:</strong> digits = &quot;2&quot;
 <strong>Output:</strong> [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;]
@@ -48,7 +41,7 @@ tags:
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>0 &lt;= digits.length &lt;= 4</code></li>
+	<li><code>1 &lt;= digits.length &lt;= 4</code></li>
 	<li><code>digits[i]</code> is a digit in the range <code>[&#39;2&#39;, &#39;9&#39;]</code>.</li>
 </ul>
 

solution/0000-0099/0026.Remove Duplicates from Sorted Array/README.md

@@ -19,12 +19,9 @@ tags:
 
 <p>给你一个 <strong>非严格递增排列</strong> 的数组 <code>nums</code> ，请你<strong><a href="http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank"> 原地</a></strong> 删除重复出现的元素，使每个元素 <strong>只出现一次</strong> ，返回删除后数组的新长度。元素的 <strong>相对顺序</strong> 应该保持 <strong>一致</strong> 。然后返回 <code>nums</code> 中唯一元素的个数。</p>
 
-<p>考虑 <code>nums</code> 的唯一元素的数量为 <code>k</code> ，你需要做以下事情确保你的题解可以被通过：</p>
+<p>考虑 <code>nums</code> 的唯一元素的数量为 <code>k</code>。去重后，返回唯一元素的数量 <code>k</code>。</p>
 
-<ul>
-	<li>更改数组 <code>nums</code> ，使 <code>nums</code> 的前 <code>k</code> 个元素包含唯一元素，并按照它们最初在 <code>nums</code> 中出现的顺序排列。<code>nums</code>&nbsp;的其余元素与 <code>nums</code> 的大小不重要。</li>
-	<li>返回 <code>k</code>&nbsp;。</li>
-</ul>
+<p><code>nums</code> 的前 <code>k</code> 个元素应包含 <strong>排序后</strong> 的唯一数字。下标&nbsp;<code>k - 1</code> 之后的剩余元素可以忽略。</p>
 
 <p><strong>判题标准:</strong></p>
 
@@ -57,7 +54,7 @@ for (int i = 0; i &lt; k; i++) {
 
 <pre>
 <strong>输入：</strong>nums = [0,0,1,1,1,2,2,3,3,4]
-<strong>输出：</strong>5, nums = [0,1,2,3,4]
+<strong>输出：</strong>5, nums = [0,1,2,3,4,_,_,_,_,_]
 <strong>解释：</strong>函数应该返回新的长度 <strong><code>5</code></strong> ， 并且原数组 <em>nums </em>的前五个元素被修改为 <strong><code>0</code></strong>, <strong><code>1</code></strong>, <strong><code>2</code></strong>, <strong><code>3</code></strong>, <strong><code>4</code></strong> 。不需要考虑数组中超出新长度后面的元素。
 </pre>
 
@@ -67,8 +64,8 @@ for (int i = 0; i &lt; k; i++) {
 
 <ul>
 	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
-	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
-	<li><code>nums</code> 已按 <strong>非严格递增</strong>&nbsp;排列</li>
+	<li><code>-10<font size="1">0</font>&nbsp;&lt;= nums[i] &lt;= 10<font size="1">0</font></code></li>
+	<li><code>nums</code> 已按 <strong>非递减</strong>&nbsp;顺序排列。</li>
 </ul>
 
 <!-- description:end -->

solution/0000-0099/0026.Remove Duplicates from Sorted Array/README_EN.md

@@ -17,14 +17,11 @@ tags:
 
 <!-- description:start -->
 
-<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing order</strong>, remove the duplicates <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank"><strong>in-place</strong></a> such that each unique element appears only <strong>once</strong>. The <strong>relative order</strong> of the elements should be kept the <strong>same</strong>. Then return <em>the number of unique elements in </em><code>nums</code>.</p>
+<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing order</strong>, remove the duplicates <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank"><strong>in-place</strong></a> such that each unique element appears only <strong>once</strong>. The <strong>relative order</strong> of the elements should be kept the <strong>same</strong>.</p>
 
-<p>Consider the number of unique elements of <code>nums</code> to be <code>k</code>, to get accepted, you need to do the following things:</p>
+<p>Consider the number of <em>unique elements</em> in&nbsp;<code>nums</code> to be <code>k<strong>​​​​​​​</strong></code>​​​​​​​. <meta charset="UTF-8" />After removing duplicates, return the number of unique elements&nbsp;<code>k</code>.</p>
 
-<ul>
-	<li>Change the array <code>nums</code> such that the first <code>k</code> elements of <code>nums</code> contain the unique elements in the order they were present in <code>nums</code> initially. The remaining elements of <code>nums</code> are not important as well as the size of <code>nums</code>.</li>
-	<li>Return <code>k</code>.</li>
-</ul>
+<p><meta charset="UTF-8" />The first&nbsp;<code>k</code>&nbsp;elements of&nbsp;<code>nums</code>&nbsp;should contain the unique numbers in <strong>sorted order</strong>. The remaining elements beyond index&nbsp;<code>k - 1</code>&nbsp;can be ignored.</p>
 
 <p><strong>Custom Judge:</strong></p>
 

solution/0000-0099/0085.Maximal Rectangle/README.md

@@ -53,7 +53,7 @@ tags:
 <ul>
 	<li><code>rows == matrix.length</code></li>
 	<li><code>cols == matrix[0].length</code></li>
-	<li><code>1 &lt;= row, cols &lt;= 200</code></li>
+	<li><code>1 &lt;= rows, cols &lt;= 200</code></li>
 	<li><code>matrix[i][j]</code> 为 <code>'0'</code> 或 <code>'1'</code></li>
 </ul>
 

solution/0000-0099/0085.Maximal Rectangle/README_EN.md

@@ -51,7 +51,7 @@ tags:
 <ul>
 	<li><code>rows == matrix.length</code></li>
 	<li><code>cols == matrix[i].length</code></li>
-	<li><code>1 &lt;= row, cols &lt;= 200</code></li>
+	<li><code>1 &lt;= rows, cols &lt;= 200</code></li>
 	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
 </ul>
 

solution/0100-0199/0118.Pascal's Triangle/README.md

@@ -17,13 +17,13 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一个非负整数 <em><code>numRows</code>，</em>生成「杨辉三角」的前 <em><code>numRows</code> </em>行。</p>
+<p>给定一个非负整数&nbsp;<em><code>numRows</code>，</em>生成「杨辉三角」的前&nbsp;<em><code>numRows</code>&nbsp;</em>行。</p>
 
-<p><small>在「杨辉三角」中，每个数是它左上方和右上方的数的和。</small></p>
+<p>在<strong>「杨辉三角」</strong>中，每个数是它左上方和右上方的数的和。</p>
 
 <p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0118.Pascal%27s%20Triangle/images/1626927345-DZmfxB-PascalTriangleAnimated2.gif" /></p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1:</strong></p>
 
@@ -32,19 +32,19 @@ tags:
 <strong>输出:</strong> [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
 </pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong>示例&nbsp;2:</strong></p>
 
 <pre>
 <strong>输入:</strong> numRows = 1
 <strong>输出:</strong> [[1]]
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示:</strong></p>
 
 <ul>
-	<li><code>1 <= numRows <= 30</code></li>
+	<li><code>1 &lt;= numRows &lt;= 30</code></li>
 </ul>
 
 <!-- description:end -->

solution/0100-0199/0122.Best Time to Buy and Sell Stock II/README_EN.md

@@ -20,7 +20,7 @@ tags:
 
 <p>You are given an integer array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day.</p>
 
-<p>On each day, you may decide to buy and/or sell the stock. You can only hold <strong>at most one</strong> share of the stock at any time. However, you can sell and buy the stock multiple times on the <strong>same day</strong>, ensuring you never hold than one share of the stock.</p>
+<p>On each day, you may decide to buy and/or sell the stock. You can only hold <strong>at most one</strong> share of the stock at any time. However, you can sell and buy the stock multiple times on the <strong>same day</strong>, ensuring you never hold more than one share of the stock.</p>
 
 <p>Find and return <em>the <strong>maximum</strong> profit you can achieve</em>.</p>
 

solution/0100-0199/0152.Maximum Product Subarray/README.md

@@ -21,6 +21,8 @@ tags:
 
 <p>测试用例的答案是一个&nbsp;<strong>32-位</strong> 整数。</p>
 
+<p><strong>请注意</strong>，一个只包含一个元素的数组的乘积是这个元素的值。</p>
+
 <p>&nbsp;</p>
 
 <p><strong class="example">示例 1:</strong></p>

solution/0100-0199/0152.Maximum Product Subarray/README_EN.md

@@ -21,6 +21,8 @@ tags:
 
 <p>The test cases are generated so that the answer will fit in a <strong>32-bit</strong> integer.</p>
 
+<p><strong>Note</strong> that the product of an array with a single element is the value of that element.</p>
+
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 

solution/0100-0199/0166.Fraction to Recurring Decimal/README.md

@@ -26,6 +26,8 @@ tags:
 
 <p>对于所有给定的输入，<strong>保证</strong> 答案字符串的长度小于 <code>10<sup>4</sup></code> 。</p>
 
+<p><strong>注意</strong>，如果分数可以表示为有限长度的字符串，则 <strong>必须</strong> 返回它。</p>
+
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>

solution/0200-0299/0218.The Skyline Problem/README.md

@@ -8,6 +8,7 @@ tags:
     - 数组
     - 分治
     - 有序集合
+    - 排序
     - 扫描线
     - 堆（优先队列）
 ---

solution/0200-0299/0218.The Skyline Problem/README_EN.md

@@ -8,6 +8,7 @@ tags:
     - Array
     - Divide and Conquer
     - Ordered Set
+    - Sorting
     - Line Sweep
     - Heap (Priority Queue)
 ---

solution/0300-0399/0352.Data Stream as Disjoint Intervals/README.md

@@ -3,8 +3,11 @@ comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0352.Data%20Stream%20as%20Disjoint%20Intervals/README.md
 tags:
+    - 并查集
     - 设计
+    - 哈希表
     - 二分查找
+    - 数据流
     - 有序集合
 ---
 

solution/0300-0399/0352.Data Stream as Disjoint Intervals/README_EN.md

@@ -3,8 +3,11 @@ comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0352.Data%20Stream%20as%20Disjoint%20Intervals/README_EN.md
 tags:
+    - Union Find
     - Design
+    - Hash Table
     - Binary Search
+    - Data Stream
     - Ordered Set
 ---
 

solution/0400-0499/0452.Minimum Number of Arrows to Burst Balloons/README.md

@@ -20,7 +20,7 @@ tags:
 
 <p>有一些球形气球贴在一堵用 XY 平面表示的墙面上。墙面上的气球记录在整数数组&nbsp;<code>points</code>&nbsp;，其中<code>points[i] = [x<sub>start</sub>, x<sub>end</sub>]</code>&nbsp;表示水平直径在&nbsp;<code>x<sub>start</sub></code>&nbsp;和&nbsp;<code>x<sub>end</sub></code>之间的气球。你不知道气球的确切 y 坐标。</p>
 
-<p>一支弓箭可以沿着 x 轴从不同点 <strong>完全垂直</strong> 地射出。在坐标 <code>x</code> 处射出一支箭，若有一个气球的直径的开始和结束坐标为 <code>x</code><sub><code>start</code>，</sub><code>x</code><sub><code>end</code>，</sub> 且满足 &nbsp;<code>x<sub>start</sub>&nbsp;≤ x ≤ x</code><sub><code>end</code>，</sub>则该气球会被 <strong>引爆</strong>&nbsp;<sub>。</sub>可以射出的弓箭的数量 <strong>没有限制</strong> 。 弓箭一旦被射出之后，可以无限地前进。</p>
+<p>一支弓箭可以沿着 x 轴从不同点 <strong>完全垂直</strong> 地射出。在坐标 <code>x</code> 处射出一支箭，若有一个气球的直径的开始和结束坐标为 <code>x<sub>start</sub></code><sub>，</sub><code>x<sub>end</sub></code><sub>，</sub> 且满足 &nbsp;<code>x<sub>start</sub>&nbsp;≤ x ≤ x<sub>end</sub></code><sub>，</sub>则该气球会被 <strong>引爆</strong>&nbsp;<sub>。</sub>可以射出的弓箭的数量 <strong>没有限制</strong> 。 弓箭一旦被射出之后，可以无限地前进。</p>
 
 <p>给你一个数组 <code>points</code> ，<em>返回引爆所有气球所必须射出的 <strong>最小</strong> 弓箭数&nbsp;</em>。</p>
 &nbsp;

solution/0400-0499/0457.Circular Array Loop/README.md

@@ -21,8 +21,8 @@ tags:
 <p>存在一个不含 <code>0</code> 的<strong> 环形 </strong>数组&nbsp;<code>nums</code> ，每个 <code>nums[i]</code> 都表示位于下标 <code>i</code> 的角色应该向前或向后移动的下标个数：</p>
 
 <ul>
-	<li>如果 <code>nums[i]</code> 是正数，<strong>向前</strong>（下标递增方向）移动 <code>|nums[i]|</code> 步</li>
-	<li>如果&nbsp;<code>nums[i]</code> 是负数，<strong>向后</strong>（下标递减方向）移动 <code>|nums[i]|</code> 步</li>
+	<li>如果 <code>nums[i]</code> 是正数，<strong>向前</strong>（下标递增方向）移动 <code>nums[i]</code>&nbsp;步</li>
+	<li>如果&nbsp;<code>nums[i]</code> 是负数，<strong>向后</strong>（下标递减方向）移动 <code>abs(nums[i])</code>&nbsp;步</li>
 </ul>
 
 <p>因为数组是 <strong>环形</strong> 的，所以可以假设从最后一个元素向前移动一步会到达第一个元素，而第一个元素向后移动一步会到达最后一个元素。</p>

solution/0400-0499/0457.Circular Array Loop/README_EN.md

@@ -22,7 +22,7 @@ tags:
 
 <ul>
 	<li>If <code>nums[i]</code> is positive, move <code>nums[i]</code> steps <strong>forward</strong>, and</li>
-	<li>If <code>nums[i]</code> is negative, move <code>nums[i]</code> steps <strong>backward</strong>.</li>
+	<li>If <code>nums[i]</code> is negative, move <code>abs(nums[i])</code> steps <strong>backward</strong>.</li>
 </ul>
 
 <p>Since the array is <strong>circular</strong>, you may assume that moving forward from the last element puts you on the first element, and moving backwards from the first element puts you on the last element.</p>

solution/0700-0799/0735.Asteroid Collision/README.md

@@ -47,6 +47,14 @@ tags:
 <strong>输出：</strong>[10]
 <b>解释：</b>2 和 -5 发生碰撞后剩下 -5 。10 和 -5 发生碰撞后剩下 10 。</pre>
 
+<p><strong class="example">示例 4：</strong></p>
+
+<pre>
+<b>输入：</b>asteroids = [3,5,-6,2,-1,4]
+<b>输出：</b>[-6,2,4]
+<b>解释：</b>小行星 -6 使小行星 3 和 5 爆炸，然后继续向左移动。在另一边，小行星 2 使小行星 -1 爆炸，然后继续向右移动，没有碰撞小行星 4。
+</pre>
+
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>

solution/0700-0799/0735.Asteroid Collision/README_EN.md

@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row. The indices of the asteriod in the array represent their relative position in space.</p>
+<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row. The indices of the asteroid in the array represent their relative position in space.</p>
 
 <p>For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.</p>
 
@@ -49,6 +49,14 @@ tags:
 <strong>Explanation:</strong> The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.
 </pre>
 
+<p><strong class="example">Example 4:</strong></p>
+
+<pre>
+<strong>Input:</strong> asteroids = [3,5,-6,2,-1,4]​​​​​​​
+<strong>Output:</strong> [-6,2,4]
+<strong>Explanation:</strong> The asteroid -6 makes the asteroid 3 and 5 explode, and then continues going left. On the other side, the asteroid 2 makes the asteroid -1 explode and then continues going right, without reaching asteroid 4.
+</pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

solution/0700-0799/0756.Pyramid Transition Matrix/README.md

@@ -4,8 +4,9 @@ difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0756.Pyramid%20Transition%20Matrix/README.md
 tags:
     - 位运算
-    - 深度优先搜索
-    - 广度优先搜索
+    - 哈希表
+    - 字符串
+    - 回溯
 ---
 
 <!-- problem:start -->

solution/0700-0799/0756.Pyramid Transition Matrix/README_EN.md

@@ -4,8 +4,9 @@ difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0756.Pyramid%20Transition%20Matrix/README_EN.md
 tags:
     - Bit Manipulation
-    - Depth-First Search
-    - Breadth-First Search
+    - Hash Table
+    - String
+    - Backtracking
 ---
 
 <!-- problem:start -->

solution/0700-0799/0765.Couples Holding Hands/README.md

@@ -22,7 +22,7 @@ tags:
 
 <p><code>n</code> 对情侣坐在连续排列的 <code>2n</code>&nbsp;个座位上，想要牵到对方的手。</p>
 
-<p>人和座位由一个整数数组 <code>row</code> 表示，其中 <code>row[i]</code> 是坐在第 <code>i </code>个座位上的人的 <strong>ID</strong>。情侣们按顺序编号，第一对是&nbsp;<code>(0, 1)</code>，第二对是&nbsp;<code>(2, 3)</code>，以此类推，最后一对是&nbsp;<code>(2n-2, 2n-1)</code>。</p>
+<p>人和座位由一个整数数组 <code>row</code> 表示，其中 <code>row[i]</code> 是坐在第 <code>i </code>个座位上的人的 <strong>ID</strong>。情侣们按顺序编号，第一对是&nbsp;<code>(0, 1)</code>，第二对是&nbsp;<code>(2, 3)</code>，以此类推，最后一对是&nbsp;<code>(2n - 2, 2n - 1)</code>。</p>
 
 <p>返回 <em>最少交换座位的次数，以便每对情侣可以并肩坐在一起</em>。 <i>每次</i>交换可选择任意两人，让他们站起来交换座位。</p>
 
@@ -33,7 +33,7 @@ tags:
 <pre>
 <strong>输入:</strong> row = [0,2,1,3]
 <strong>输出:</strong> 1
-<strong>解释:</strong> 只需要交换row[1]和row[2]的位置即可。
+<strong>解释:</strong> 只需要交换第二个人（row[1]）和第三个人（row[2]）的位置即可。
 </pre>
 
 <p><strong>示例 2:</strong></p>
@@ -53,7 +53,7 @@ tags:
 	<li><code>2 &lt;= n &lt;= 30</code></li>
 	<li><code>n</code>&nbsp;是偶数</li>
 	<li><code>0 &lt;= row[i] &lt; 2n</code></li>
-	<li><code>row</code>&nbsp;中所有元素均<strong>无重复</strong></li>
+	<li><code>row</code>&nbsp;中所有元素均&nbsp;<strong>无重复</strong></li>
 </ul>
 
 <!-- description:end -->

solution/0700-0799/0765.Couples Holding Hands/README_EN.md

@@ -48,8 +48,7 @@ tags:
 
 <ul>
 	<li><code>2n == row.length</code></li>
-	<li><code>2 &lt;= n &lt;= 30</code></li>
-	<li><code>n</code> is even.</li>
+	<li><code>2 &lt;= n &lt;= 30</code>​​​​​​​</li>
 	<li><code>0 &lt;= row[i] &lt; 2n</code></li>
 	<li>All the elements of <code>row</code> are <strong>unique</strong>.</li>
 </ul>

solution/0900-0999/0948.Bag of Tokens/README.md

@@ -40,7 +40,7 @@ tags:
 <pre>
 <strong>输入：</strong>tokens = [100], power = 50
 <strong>输出：</strong>0
-<strong>解释：</strong>因为你的初始分数为 <code>0，</code>无法使令牌朝下。你也不能使令牌朝上因为你的能量（<code>50</code>）比 <code>tokens[0]</code>&nbsp;少（<code>100</code>）。</pre>
+<strong>解释：</strong>因为你的初始分数为 <code>0</code>，无法使令牌朝下。你也不能使令牌朝上因为你的能量（<code>50</code>）比 <code>tokens[0]</code>&nbsp;少（<code>100</code>）。</pre>
 
 <p><strong>示例 2：</strong></p>
 

solution/1000-1099/1025.Divisor Game/README.md

@@ -26,7 +26,7 @@ tags:
 <p>最初，黑板上有一个数字&nbsp;<code>n</code>&nbsp;。在每个玩家的回合，玩家需要执行以下操作：</p>
 
 <ul>
-	<li>选出任一&nbsp;<code>x</code>，满足&nbsp;<code>0 &lt; x &lt; n</code>&nbsp;且&nbsp;<code>n % x == 0</code>&nbsp;。</li>
+	<li>选出任一整数&nbsp;<code>x</code>，满足&nbsp;<code>0 &lt; x &lt; n</code>&nbsp;且&nbsp;<code>n % x == 0</code>&nbsp;。</li>
 	<li>用 <code>n - x</code>&nbsp;替换黑板上的数字&nbsp;<code>n</code> 。</li>
 </ul>