feat: add solutions to lc problem: No.1020 (#3553)

doocs · Sep 23, 2024 · 4fab89d · 4fab89d
1 parent 44b8480
commit 4fab89d
Show file tree

Hide file tree

Showing 27 changed files with 432 additions and 770 deletions.
diff --git a/solution/0600-0699/0621.Task Scheduler/README.md b/solution/0600-0699/0621.Task Scheduler/README.md
@@ -29,33 +29,37 @@ tags:
 
 <p><strong>示例 1：</strong></p>
 
-<pre>
-<strong>输入：</strong>tasks = ["A","A","A","B","B","B"], n = 2
-<strong>输出：</strong>8
-<strong>解释：</strong>A -&gt; B -&gt; (待命) -&gt; A -&gt; B -&gt; (待命) -&gt; A -&gt; B
-     在本示例中，两个相同类型任务之间必须间隔长度为 n = 2 的冷却时间，而执行一个任务只需要一个单位时间，所以中间出现了（待命）状态。 </pre>
-
-<p><strong>示例 2：</strong></p>
-
-<pre>
-<strong>输入：</strong>tasks = ["A","A","A","B","B","B"], n = 0
-<strong>输出：</strong>6
-<strong>解释：</strong>在这种情况下，任何大小为 6 的排列都可以满足要求，因为 n = 0
-["A","A","A","B","B","B"]
-["A","B","A","B","A","B"]
-["B","B","B","A","A","A"]
-...
-诸如此类
-</pre>
+<div class="example-block"><strong>输入：</strong>tasks = ["A","A","A","B","B","B"], n = 2</div>
+
+<div class="example-block"><strong>输出：</strong>8</div>
+
+<div class="example-block"><strong>解释：</strong></div>
+
+<div class="example-block">在完成任务 A 之后，你必须等待两个间隔。对任务 B 来说也是一样。在第 3 个间隔，A 和 B 都不能完成，所以你需要待命。在第 4 个间隔，由于已经经过了 2 个间隔，你可以再次执行 A 任务。</div>
+
+<div class="example-block">&nbsp;</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><b>输入：</b>tasks = ["A","C","A","B","D","B"], n = 1</p>
+
+<p><b>输出：</b>6</p>
+
+<p><b>解释：</b>一种可能的序列是：A -&gt; B -&gt; C -&gt; D -&gt; A -&gt; B。</p>
+
+<p>由于冷却间隔为 1，你可以在完成另一个任务后重复执行这个任务。</p>
+</div>
 
 <p><strong>示例 3：</strong></p>
 
-<pre>
-<strong>输入：</strong>tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
-<strong>输出：</strong>16
-<strong>解释：</strong>一种可能的解决方案是：
-     A -&gt; B -&gt; C -&gt; A -&gt; D -&gt; E -&gt; A -&gt; F -&gt; G -&gt; A -&gt; (待命) -&gt; (待命) -&gt; A -&gt; (待命) -&gt; (待命) -&gt; A
-</pre>
+<div class="example-block"><strong>输入：</strong>tasks = ["A","A","A","B","B","B"], n = 0</div>
+
+<div class="example-block"><strong>输出：</strong>6</div>
+
+<div class="example-block"><strong>解释：</strong>一种可能的序列为：A -&gt; B -&gt; idle -&gt; idle -&gt; A -&gt; B -&gt; idle -&gt; idle -&gt; A -&gt; B。</div>
+
+<div class="example-block">只有两种任务类型，A 和 B，需要被 3 个间隔分割。这导致重复执行这些任务的间隔当中有两次待命状态。</div>
 
 <p>&nbsp;</p>
 

diff --git a/solution/0600-0699/0621.Task Scheduler/README_EN.md b/solution/0600-0699/0621.Task Scheduler/README_EN.md
@@ -21,9 +21,9 @@ tags:
 
 <!-- description:start -->
 
-<p>You are given an array of CPU <code>tasks</code>, each represented by letters&nbsp;A&nbsp;to Z, and a cooling time, <code>n</code>. Each cycle or interval allows the completion of one task. Tasks can be completed in any order, but there&#39;s a constraint: <strong>identical</strong> tasks must be separated by at least <code>n</code> intervals due to cooling time.</p>
+<p>You are given an array of CPU <code>tasks</code>, each labeled with a letter from A to Z, and a number <code>n</code>. Each CPU interval can be idle or allow the completion of one task. Tasks can be completed in any order, but there&#39;s a constraint: there has to be a gap of <strong>at least</strong> <code>n</code> intervals between two tasks with the same label.</p>
 
-<p>Return the <em>minimum number of intervals</em> required to complete all tasks.</p>
+<p>Return the <strong>minimum</strong> number of CPU intervals required to complete all tasks.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
@@ -50,7 +50,7 @@ font-size: 0.85rem;
 
 <p><strong>Explanation:</strong> A possible sequence is: A -&gt; B -&gt; idle -&gt; A -&gt; B -&gt; idle -&gt; A -&gt; B.</p>
 
-<p>After completing task A, you must wait two cycles before doing A again. The same applies to task B. In the 3<sup>rd</sup> interval, neither A nor B can be done, so you idle. By the 4<sup>th</sup> cycle, you can do A again as 2 intervals have passed.</p>
+<p>After completing task A, you must wait two intervals before doing A again. The same applies to task B. In the 3<sup>rd</sup> interval, neither A nor B can be done, so you idle. By the 4<sup>th</sup> interval, you can do A again as 2 intervals have passed.</p>
 </div>
 
 <p><strong class="example">Example 2:</strong></p>

diff --git a/solution/0600-0699/0658.Find K Closest Elements/README.md b/solution/0600-0699/0658.Find K Closest Elements/README.md
@@ -42,8 +42,8 @@ tags:
 <p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>输入：</strong>arr = [1,2,3,4,5], k = 4, x = -1
-<strong>输出：</strong>[1,2,3,4]
+<strong>输入：</strong>arr = [1,1,2,3,4,5], k = 4, x = -1
+<strong>输出：</strong>[1,1,2,3]
 </pre>
 
 <p>&nbsp;</p>