docs: code format

Author: yanglbme

basic/searching/BinarySearch/README.md

@@ -25,9 +25,12 @@ int search(int left, int right) {
 ### 模板 2
 
 ```java
-boolean check(int x) {}
+boolean
+check(int x) {
+}
 
-int search(int left, int right) {
+int
+search(int left, int right) {
     while (left < right) {
         int mid = (left + right + 1) >> 1;
         if (check(mid)) {

basic/searching/BinarySearch/README_EN.md

@@ -23,9 +23,12 @@ int search(int left, int right) {
 ### Template 2
 
 ```java
-boolean check(int x) {}
+boolean
+check(int x) {
+}
 
-int search(int left, int right) {
+int
+search(int left, int right) {
     while (left < right) {
         int mid = (left + right + 1) >> 1;
         if (check(mid)) {

basic/sorting/CountingSort/README.md

@@ -17,7 +17,8 @@
 ### **Java**
 
 ```java
-public static void sort(int[] nums, int min, int max) {
+public static void
+sort(int[] nums, int min, int max) {
     int[] c = new int[max - min + 1];
     for (int v : nums) {
         c[v - min]++;

basic/sorting/InsertionSort/README.md

@@ -27,7 +27,7 @@ public class InsertionSort {
     private static void insertionSort(int[] nums) {
         for (int i = 1, j, n = nums.length; i < n; ++i) {
             int num = nums[i];
-            for (j = i - 1; j >=0 && nums[j] > num; --j) {
+            for (j = i - 1; j >= 0 && nums[j] > num; --j) {
                 nums[j + 1] = nums[j];
             }
             nums[j + 1] = num;

basic/sorting/QuickSort/README.md

@@ -121,8 +121,10 @@ public class Main {
         int i = left - 1, j = right + 1;
         int x = nums[(left + right) >> 1];
         while (i < j) {
-            while (nums[++i] < x);
-            while (nums[--j] > x);
+            while (nums[++i] < x)
+                ;
+            while (nums[--j] > x)
+                ;
             if (i < j) {
                 int t = nums[i];
                 nums[i] = nums[j];

basic/sorting/SelectionSort/README.md

@@ -37,7 +37,6 @@ public class SelectionSort {
         System.out.println(Arrays.toString(nums));
     }
 }
-
 ```
 
 ### **JavaScript**

basic/sorting/ShellSort/README.md

@@ -36,7 +36,7 @@ public class ShellSort {
     }
 
     public static void main(String[] args) {
-        System.out.println(Arrays.toString(shellSort(new int[]{1, 2, 7, 9, 5, 8})));
+        System.out.println(Arrays.toString(shellSort(new int[] {1, 2, 7, 9, 5, 8})));
     }
 }
 ```

lcci/02.05.Sum Lists/README.md

@@ -80,8 +80,7 @@ class Solution {
         ListNode dummy = new ListNode(-1);
         ListNode cur = dummy;
         while (l1 != null || l2 != null || carry != 0) {
-            int s =
-                (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
+            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
             carry = s / 10;
             cur.next = new ListNode(s % 10);
             cur = cur.next;

lcci/02.05.Sum Lists/README_EN.md

@@ -75,8 +75,7 @@ class Solution {
         ListNode dummy = new ListNode(-1);
         ListNode cur = dummy;
         while (l1 != null || l2 != null || carry != 0) {
-            int s =
-                (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
+            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
             carry = s / 10;
             cur.next = new ListNode(s % 10);
             cur = cur.next;

lcci/03.03.Stack of Plates/README.md

@@ -1,68 +1,45 @@
 # [面试题 03.03. 堆盘子](https://leetcode.cn/problems/stack-of-plates-lcci)
-
 [English Version](/lcci/03.03.Stack%20of%20Plates/README_EN.md)
-
 ## 题目描述
-
 <!-- 这里写题目描述 -->
 <p>堆盘子。设想有一堆盘子，堆太高可能会倒下来。因此，在现实生活中，盘子堆到一定高度时，我们就会另外堆一堆盘子。请实现数据结构<code>SetOfStacks</code>，模拟这种行为。<code>SetOfStacks</code>应该由多个栈组成，并且在前一个栈填满时新建一个栈。此外，<code>SetOfStacks.push()</code>和<code>SetOfStacks.pop()</code>应该与普通栈的操作方法相同（也就是说，pop()返回的值，应该跟只有一个栈时的情况一样）。 进阶：实现一个<code>popAt(int index)</code>方法，根据指定的子栈，执行pop操作。</p>
-
 <p>当某个栈为空时，应当删除该栈。当栈中没有元素或不存在该栈时，<code>pop</code>，<code>popAt</code>&nbsp;应返回 -1.</p>
-
 <p><strong>示例1:</strong></p>
-
 <pre><strong> 输入</strong>：
 [&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;pop&quot;, &quot;pop&quot;]
 [[1], [1], [2], [1], [], []]
 <strong> 输出</strong>：
 [null, null, null, 2, 1, -1]
 </pre>
-
 <p><strong>示例2:</strong></p>
-
 <pre><strong> 输入</strong>：
 [&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;popAt&quot;, &quot;popAt&quot;]
 [[2], [1], [2], [3], [0], [0], [0]]
 <strong> 输出</strong>：
 [null, null, null, null, 2, 1, 3]
 </pre>
-
 ## 解法
-
 <!-- 这里可写通用的实现逻辑 -->
-
 <!-- tabs:start -->
-
 ### **Python3**
-
 <!-- 这里可写当前语言的特殊实现逻辑 -->
-
 ```python
 
-
 ```
-
 ### **Java**
-
 <!-- 这里可写当前语言的特殊实现逻辑 -->
-
 ```java
 
-
 ```
-
 ### **TypeScript**
-
 ```ts
 class StackOfPlates {
     private cap: number;
     private stacks: number[][];
-
     constructor(cap: number) {
         this.cap = cap;
         this.stacks = [];
     }
-
     push(val: number): void {
         if (this.cap === 0) {
             return;
@@ -75,7 +52,6 @@ class StackOfPlates {
             stack.push(val);
         }
     }
-
     pop(): number {
         const n = this.stacks.length;
         if (n === 0) {
@@ -88,7 +64,6 @@ class StackOfPlates {
         }
         return res;
     }
-
     popAt(index: number): number {
         if (index >= this.stacks.length) {
             return -1;
@@ -101,7 +76,6 @@ class StackOfPlates {
         return res;
     }
 }
-
 /**
  * Your StackOfPlates object will be instantiated and called as such:
  * var obj = new StackOfPlates(cap)
@@ -110,12 +84,8 @@ class StackOfPlates {
  * var param_3 = obj.popAt(index)
  */
 ```
-
 ### **...**
-
 ```
 
-
 ```
-
 <!-- tabs:end -->

lcci/03.03.Stack of Plates/README_EN.md

@@ -1,103 +1,57 @@
 # [03.03. Stack of Plates](https://leetcode.cn/problems/stack-of-plates-lcci)
-
 [中文文档](/lcci/03.03.Stack%20of%20Plates/README.md)
-
 ## Description
-
 <p>Imagine a (literal) stack of plates. If the stack gets too high, it might topple. Therefore, in real life, we would likely start a new stack when the previous stack exceeds some threshold. Implement a data structure <code>SetOfStacks</code> that mimics this.&nbsp;<code>SetOfStacks</code> should be composed of several stacks and should create a new stack once the previous one exceeds capacity. <code>SetOfStacks.push()</code> and <code>SetOfStacks.pop()</code> should behave identically to a single stack (that is, <code>pop()</code> should return the same values as it would if there were just a single stack). Follow Up: Implement a function <code>popAt(int index)</code> which performs a pop operation on a specific sub-stack.</p>
-
 <p>You should delete the sub-stack when it becomes empty. <code>pop</code>, <code>popAt</code> should return -1 when there&#39;s no element to pop.</p>
-
 <p><strong>Example1:</strong></p>
-
 <pre>
 
-
-
 <strong> Input</strong>: 
 
-
-
 [&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;pop&quot;, &quot;pop&quot;]
 
-
-
 [[1], [1], [2], [1], [], []]
 
-
-
 <strong> Output</strong>: 
 
-
-
 [null, null, null, 2, 1, -1]
 
-
-
 <strong> Explanation</strong>: 
 
-
-
 </pre>
-
 <p><strong>Example2:</strong></p>
-
 <pre>
 
-
-
 <strong> Input</strong>: 
 
-
-
 [&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;popAt&quot;, &quot;popAt&quot;]
 
-
-
 [[2], [1], [2], [3], [0], [0], [0]]
 
-
-
 <strong> Output</strong>: 
 
-
-
 [null, null, null, null, 2, 1, 3]
 
-
-
 </pre>
-
 ## Solutions
-
 <!-- tabs:start -->
-
 ### **Python3**
-
 ```python
 
-
 ```
-
 ### **Java**
-
 ```java
 
-
 ```
-
 ### **TypeScript**
-
 ```ts
 class StackOfPlates {
     private cap: number;
     private stacks: number[][];
-
     constructor(cap: number) {
         this.cap = cap;
         this.stacks = [];
     }
-
     push(val: number): void {
         if (this.cap === 0) {
             return;
@@ -110,7 +64,6 @@ class StackOfPlates {
             stack.push(val);
         }
     }
-
     pop(): number {
         const n = this.stacks.length;
         if (n === 0) {
@@ -123,7 +76,6 @@ class StackOfPlates {
         }
         return res;
     }
-
     popAt(index: number): number {
         if (index >= this.stacks.length) {
             return -1;
@@ -136,7 +88,6 @@ class StackOfPlates {
         return res;
     }
 }
-
 /**
  * Your StackOfPlates object will be instantiated and called as such:
  * var obj = new StackOfPlates(cap)
@@ -145,12 +96,8 @@ class StackOfPlates {
  * var param_3 = obj.popAt(index)
  */
 ```
-
 ### **...**
-
 ```
 
-
 ```
-
 <!-- tabs:end -->

lcci/03.06.Animal Shelter/README.md

@@ -103,15 +103,18 @@ class AnimalShelf {
     }
 
     public int[] dequeueAny() {
-        return dogs.isEmpty() ? dequeueCat() : (cats.isEmpty() ? dequeueDog() : (dogs.peek() < cats.peek() ? dequeueDog() : dequeueCat()));
+        return dogs.isEmpty()
+            ? dequeueCat()
+            : (cats.isEmpty() ? dequeueDog()
+                              : (dogs.peek() < cats.peek() ? dequeueDog() : dequeueCat()));
     }
 
     public int[] dequeueDog() {
-        return dogs.isEmpty() ? new int[]{-1, -1} : new int[]{dogs.poll(), 1};
+        return dogs.isEmpty() ? new int[] {-1, -1} : new int[] {dogs.poll(), 1};
     }
 
     public int[] dequeueCat() {
-        return cats.isEmpty() ? new int[]{-1, -1} : new int[]{cats.poll(), 0};
+        return cats.isEmpty() ? new int[] {-1, -1} : new int[] {cats.poll(), 0};
     }
 }
 

lcci/03.06.Animal Shelter/README_EN.md

@@ -108,15 +108,18 @@ class AnimalShelf {
     }
 
     public int[] dequeueAny() {
-        return dogs.isEmpty() ? dequeueCat() : (cats.isEmpty() ? dequeueDog() : (dogs.peek() < cats.peek() ? dequeueDog() : dequeueCat()));
+        return dogs.isEmpty()
+            ? dequeueCat()
+            : (cats.isEmpty() ? dequeueDog()
+                              : (dogs.peek() < cats.peek() ? dequeueDog() : dequeueCat()));
     }
 
     public int[] dequeueDog() {
-        return dogs.isEmpty() ? new int[]{-1, -1} : new int[]{dogs.poll(), 1};
+        return dogs.isEmpty() ? new int[] {-1, -1} : new int[] {dogs.poll(), 1};
     }
 
     public int[] dequeueCat() {
-        return cats.isEmpty() ? new int[]{-1, -1} : new int[]{cats.poll(), 0};
+        return cats.isEmpty() ? new int[] {-1, -1} : new int[] {cats.poll(), 0};
     }
 }