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Commit 1c5edd5

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feat: add solutions to lc problems: No.1188,1196
* No.1188.Design Bounded Blocking Queue * No.1196.How Many Apples Can You Put into the Basket
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‎solution/0100-0199/0167.Two Sum II - Input Array Is Sorted/README_EN.md‎

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## Description
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<p>Given a <strong>1-indexed</strong> array of integers <code>numbers</code> that is already <strong><em>sorted in non-decreasing order</em></strong>, find two numbers such that they add up to a specific <code>target</code> number. Let these two numbers be <code>numbers[index<sub>1</sub>]</code> and <code>numbers[index<sub>2</sub>]</code> where <code>1 &lt;= index<sub>1</sub> &lt; index<sub>2</sub> &lt;= numbers.length</code>.</p>
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<p>Given a <strong>1-indexed</strong> array of integers <code>numbers</code> that is already <strong><em>sorted in non-decreasing order</em></strong>, find two numbers such that they add up to a specific <code>target</code> number. Let these two numbers be <code>numbers[index<sub>1</sub>]</code> and <code>numbers[index<sub>2</sub>]</code> where <code>1 &lt;= index<sub>1</sub> &lt; index<sub>2</sub> &lt;&nbsp;numbers.length</code>.</p>
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<p>Return<em> the indices of the two numbers, </em><code>index<sub>1</sub></code><em> and </em><code>index<sub>2</sub></code><em>, <strong>added by one</strong> as an integer array </em><code>[index<sub>1</sub>, index<sub>2</sub>]</code><em> of length 2.</em></p>
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‎solution/0500-0599/0594.Longest Harmonious Subsequence/README_EN.md‎

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<p>A <strong>subsequence</strong> of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,3,2,2,5,2,3,7]
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The longest harmonious subsequence is [3,2,2,2,3].
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3,4]
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<strong>Output:</strong> 2
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,1,1,1]
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<strong>Output:</strong> 0
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
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<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
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<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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## Solutions

‎solution/0600-0699/0684.Redundant Connection/README.md‎

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<!-- 这里写题目描述 -->
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<p>树可以看成是一个连通且 <strong>无环</strong>的<strong>无向</strong>图。</p>
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<p>树可以看成是一个连通且 <strong>无环&nbsp;</strong>的&nbsp;<strong>无向&nbsp;</strong>图。</p>
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<p>给定往一棵<code>n</code> 个节点 (节点值<code>1〜n</code>) 的树中添加一条边后的图。添加的边的两个顶点包含在 <code>1</code> 到 <code>n</code>中间,且这条附加的边不属于树中已存在的边。图的信息记录于长度为 <code>n</code> 的二维数组 <code>edges</code>,<code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>表示图中在 <code>ai</code> 和 <code>bi</code> 之间存在一条边。</p>
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<p>给定往一棵&nbsp;<code>n</code> 个节点 (节点值&nbsp;<code>1〜n</code>) 的树中添加一条边后的图。添加的边的两个顶点包含在 <code>1</code> 到 <code>n</code>&nbsp;中间,且这条附加的边不属于树中已存在的边。图的信息记录于长度为 <code>n</code> 的二维数组 <code>edges</code>&nbsp;,<code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>&nbsp;表示图中在 <code>ai</code> 和 <code>bi</code> 之间存在一条边。</p>
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<p>请找出一条可以删去的边,删除后可使得剩余部分是一个有着 <code>n</code> 个节点的树。如果有多个答案,则返回数组<code>edges</code> 中最后出现的边。</p>
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<p>请找出一条可以删去的边,删除后可使得剩余部分是一个有着 <code>n</code> 个节点的树。如果有多个答案,则返回数组&nbsp;<code>edges</code>&nbsp;中最后出现的那个。</p>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<strong>输出:</strong> [1,4]
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</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>n == edges.length</code></li>
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<li><code>3 <= n <= 1000</code></li>
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<li><code>3 &lt;= n &lt;= 1000</code></li>
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<li><code>edges[i].length == 2</code></li>
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<li><code>1 <= ai < bi <= edges.length</code></li>
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<li><code>1 &lt;= ai&nbsp;&lt; bi&nbsp;&lt;= edges.length</code></li>
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<li><code>ai != bi</code></li>
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<li><code>edges</code> 中无重复元素</li>
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<li>给定的图是连通的</li>
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<li>给定的图是连通的&nbsp;</li>
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</ul>
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## 解法

‎solution/0800-0899/0827.Making A Large Island/README_EN.md‎

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<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> grid = [[1,0],[0,1]]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> grid = [[1,1],[1,0]]
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<strong>Output:</strong> 4
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<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> grid = [[1,1],[1,1]]
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>n == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 &lt;= n &lt;= 500</code></li>
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<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
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<li><code>n == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 &lt;= n &lt;= 500</code></li>
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<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
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</ul>
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## Solutions

‎solution/1100-1199/1188.Design Bounded Blocking Queue/README.md‎

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<!-- tabs:start -->
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### **SQL**
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### **Python3**
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```sql
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```python
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from threading import Semaphore
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class BoundedBlockingQueue(object):
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def __init__(self, capacity: int):
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self.s1 = Semaphore(capacity)
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self.s2 = Semaphore(0)
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self.q = deque()
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def enqueue(self, element: int) -> None:
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self.s1.acquire()
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self.q.append(element)
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self.s2.release()
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def dequeue(self) -> int:
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self.s2.acquire()
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ans = self.q.popleft()
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self.s1.release()
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return ans
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def size(self) -> int:
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return len(self.q)
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```
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### **Java**
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```java
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class BoundedBlockingQueue {
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private Semaphore s1;
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private Semaphore s2;
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private Deque<Integer> q = new ArrayDeque<>();
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public BoundedBlockingQueue(int capacity) {
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s1 = new Semaphore(capacity);
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s2 = new Semaphore(0);
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}
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public void enqueue(int element) throws InterruptedException {
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s1.acquire();
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q.offer(element);
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s2.release();
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}
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public int dequeue() throws InterruptedException {
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s2.acquire();;
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int ans = q.poll();
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s1.release();
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return ans;
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}
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public int size() {
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return q.size();
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}
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}
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```
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### **C++**
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```cpp
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#include <semaphore.h>
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class BoundedBlockingQueue {
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public:
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BoundedBlockingQueue(int capacity) {
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sem_init(&s1, 0, capacity);
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sem_init(&s2, 0, 0);
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}
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void enqueue(int element) {
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sem_wait(&s1);
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q.push(element);
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sem_post(&s2);
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}
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int dequeue() {
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sem_wait(&s2);
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int ans = q.front();
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q.pop();
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sem_post(&s1);
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return ans;
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}
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int size() {
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return q.size();
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}
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private:
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queue<int> q;
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sem_t s1, s2;
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};
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```
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### \*\*
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<!-- tabs:end -->

‎solution/1100-1199/1188.Design Bounded Blocking Queue/README_EN.md‎

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<!-- tabs:start -->
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### **SQL**
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### **Python3**
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```sql
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```python
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from threading import Semaphore
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class BoundedBlockingQueue(object):
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def __init__(self, capacity: int):
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self.s1 = Semaphore(capacity)
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self.s2 = Semaphore(0)
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self.q = deque()
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def enqueue(self, element: int) -> None:
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self.s1.acquire()
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self.q.append(element)
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self.s2.release()
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def dequeue(self) -> int:
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self.s2.acquire()
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ans = self.q.popleft()
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self.s1.release()
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return ans
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def size(self) -> int:
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return len(self.q)
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```
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### **Java**
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```java
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class BoundedBlockingQueue {
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private Semaphore s1;
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private Semaphore s2;
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private Deque<Integer> q = new ArrayDeque<>();
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public BoundedBlockingQueue(int capacity) {
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s1 = new Semaphore(capacity);
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s2 = new Semaphore(0);
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}
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public void enqueue(int element) throws InterruptedException {
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s1.acquire();
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q.offer(element);
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s2.release();
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}
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public int dequeue() throws InterruptedException {
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s2.acquire();;
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int ans = q.poll();
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s1.release();
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return ans;
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}
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public int size() {
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return q.size();
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}
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}
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```
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### **C++**
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```cpp
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#include <semaphore.h>
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class BoundedBlockingQueue {
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public:
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BoundedBlockingQueue(int capacity) {
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sem_init(&s1, 0, capacity);
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sem_init(&s2, 0, 0);
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}
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void enqueue(int element) {
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sem_wait(&s1);
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q.push(element);
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sem_post(&s2);
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}
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int dequeue() {
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sem_wait(&s2);
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int ans = q.front();
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q.pop();
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sem_post(&s1);
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return ans;
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}
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int size() {
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return q.size();
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}
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private:
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queue<int> q;
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sem_t s1, s2;
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};
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```
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<!-- tabs:end -->

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