feat: add solutions to lc problem: No.2783 (doocs#1259)

No.2783.Flight Occupancy and Waitlist Analysis

Author: yanglbme

solution/0200-0299/0253.Meeting Rooms II/README.md

@@ -134,7 +134,7 @@ impl Solution {
                 }
             }
         }
-        
+
         pq.len() as i32
     }
 }

solution/0200-0299/0253.Meeting Rooms II/README_EN.md

@@ -115,7 +115,7 @@ impl Solution {
                 }
             }
         }
-        
+
         pq.len() as i32
     }
 }

solution/1200-1299/1229.Meeting Scheduler/README.md

@@ -157,7 +157,7 @@ impl Solution {
         let mut j: usize = 0;
         let N = slots1.len();
         let M = slots2.len();
-        
+
         while i < N && j < M {
             let (start, end) = (slots1[i][0], slots1[i][1]);
             while j < M && slots2[j][0] < end {

solution/1200-1299/1229.Meeting Scheduler/README_EN.md

@@ -141,7 +141,7 @@ impl Solution {
         let mut j: usize = 0;
         let N = slots1.len();
         let M = slots2.len();
-        
+
         while i < N && j < M {
             let (start, end) = (slots1[i][0], slots1[i][1]);
             while j < M && slots2[j][0] < end {

solution/2400-2499/2448.Minimum Cost to Make Array Equal/README.md

@@ -279,7 +279,7 @@ impl Solution {
             cur_sum += cost[i] as i64;
             i += 1;
         }
-        
+
         Self::compute_manhattan_dis(&nums, &cost, nums[i])
     }
 

solution/2400-2499/2448.Minimum Cost to Make Array Equal/README_EN.md

@@ -237,7 +237,7 @@ impl Solution {
             cur_sum += cost[i] as i64;
             i += 1;
         }
-        
+
         Self::compute_manhattan_dis(&nums, &cost, nums[i])
     }
 

solution/2500-2599/2553.Separate the Digits in an Array/README.md

@@ -183,7 +183,7 @@ impl Solution {
 
         for n in nums {
             let mut t = vec![];
-            let mut x = n; 
+            let mut x = n;
 
             while x != 0 {
                 t.push(x % 10);

solution/2500-2599/2553.Separate the Digits in an Array/README_EN.md

@@ -169,7 +169,7 @@ impl Solution {
 
         for n in nums {
             let mut t = vec![];
-            let mut x = n; 
+            let mut x = n;
 
             while x != 0 {
                 t.push(x % 10);

solution/2700-2799/2726.Calculator with Method Chaining/README.md

@@ -15,7 +15,7 @@
 	<li><code>subtract</code> - 从 <code>result</code> 中减去给定的数字 <code>value</code>&nbsp;，并返回更新后的 <code>Calculator</code> 对象。</li>
 	<li><code>multiply</code> - 将 <code>result</code> 乘以给定的数字 <code>value</code> ，并返回更新后的&nbsp;<code>Calculator</code> 对象。</li>
 	<li><code>divide</code> - 将 <code>result</code> 除以给定的数字 <code>value</code> ，并返回更新后的&nbsp;<code>Calculator</code> 对象。如果传入的值为 <code>0</code> ，则抛出错误 <code>"Division by zero is not allowed"</code> 。</li>
-	<li><code>power</code> - 将 <code>result</code> 的值乘以给定的数字 <code>value</code> ，并返回更新后的&nbsp;<code>Calculator</code> 对象。</li>
+	<li><code>power</code> - 计算 <code>result</code> 的幂，指数为给定的数字 <code>value</code> ，并返回更新后的&nbsp;<code>Calculator</code> 对象。（<code>result = result ^ value</code> ）</li>
 	<li><code>getResult</code> - 返回 <code>result</code> 的值。</li>
 </ul>
 

solution/2700-2799/2773.Height of Special Binary Tree/README.md

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定一棵具有 <code>n</code> 个节点的二叉树的根节点 <code>root</code> 。二叉树的节点编号从 <code>1</code> 到 <code>n</code> 。假设这棵树有 <code>k</code> 个叶子，顺序如下：<code>b<sub>1</sub> &lt; b<sub>2</sub> &lt; ... &lt; b<sub>k</sub></code> 。</p>
+<p>给定一棵具有 <code>n</code> 个节点的 <strong>特殊</strong> 二叉树的根节点 <code>root</code> 。特殊二叉树的节点编号从 <code>1</code> 到 <code>n</code> 。假设这棵树有 <code>k</code> 个叶子，顺序如下：<code>b<sub>1</sub> &lt; b<sub>2</sub> &lt; ... &lt; b<sub>k</sub></code> 。</p>
 
 <p>这棵树的叶子节点有一个 <strong>特殊</strong> 属性 ！对于每个叶子节点 <code>b<sub>i</sub></code> ，满足以下条件：</p>