feat: add solutions to lc problems: No.0319,0323 (#2755)

* No.0319.Bulb Switcher
* No.0323.Number of Connected Components in an Undirected Graph

Author: yanglbme

solution/0300-0399/0319.Bulb Switcher/README.md

@@ -56,14 +56,28 @@
 
 ## 解法
 
-### 方法一
+### 方法一：数学
+
+我们不妨将 $n$ 个灯泡编号为 $1, 2, 3, \cdots, n$，那么对于第 $i$ 个灯泡，它会在第 $d$ 轮被操作，当且仅当 $d$ 是 $i$ 的因子。
+
+对于一个数 $i$，它的因子个数是有限的，且因子个数为奇数时，最后的状态是开启的，否则是关闭的。
+
+因此，我们只需要找到 $1$ 到 $n$ 中因子个数为奇数的数的个数即可。
+
+对于一个数 $i$，如果它有因子 $d$，那么它一定有因子 $i/d$，因此因子个数为奇数的数一定是平方数。
+
+举个例子，数字 $12$ 的因子有 $1, 2, 3, 4, 6, 12$，因子个数为 $6$，是偶数；而对于数字 $16$ 这个平方数，因子有 $1, 2, 4, 8, 16$，因子个数为 $5$，是奇数。
+
+因此，我们只需要找到 $1$ 到 $n$ 中有多少个平方数即可，即 $\lfloor \sqrt{n} \rfloor$。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def bulbSwitch(self, n: int) -> int:
-        return int(n ** (1 / 2))
+        return int(sqrt(n))
 ```
 
 ```java
@@ -74,6 +88,27 @@ class Solution {
 }
 ```
 
+```cpp
+class Solution {
+public:
+    int bulbSwitch(int n) {
+        return (int) sqrt(n);
+    }
+};
+```
+
+```go
+func bulbSwitch(n int) int {
+	return int(math.Sqrt(float64(n)))
+}
+```
+
+```ts
+function bulbSwitch(n: number): number {
+    return Math.floor(Math.sqrt(n));
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/0300-0399/0319.Bulb Switcher/README_EN.md

@@ -47,14 +47,28 @@ So you should return 1 because there is only one bulb is on.</pre>
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Mathematics
+
+We can number the $n$ bulbs as $1, 2, 3, \cdots, n$. For the $i$-th bulb, it will be operated in the $d$-th round if and only if $d$ is a factor of $i$.
+
+For a number $i$, the number of its factors is finite. If the number of factors is odd, the final state is on; otherwise, it is off.
+
+Therefore, we only need to find the number of numbers from $1$ to $n$ with an odd number of factors.
+
+For a number $i$, if it has a factor $d$, then it must have a factor $i/d$. Therefore, numbers with an odd number of factors must be perfect squares.
+
+For example, the factors of the number $12$ are $1, 2, 3, 4, 6, 12$, and the number of factors is $6$, which is even. For the perfect square number $16$, the factors are $1, 2, 4, 8, 16$, and the number of factors is $5$, which is odd.
+
+Therefore, we only need to find how many perfect squares there are from $1$ to $n$, which is $\lfloor \sqrt{n} \rfloor$.
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def bulbSwitch(self, n: int) -> int:
-        return int(n ** (1 / 2))
+        return int(sqrt(n))
 ```
 
 ```java
@@ -65,6 +79,27 @@ class Solution {
 }
 ```
 
+```cpp
+class Solution {
+public:
+    int bulbSwitch(int n) {
+        return (int) sqrt(n);
+    }
+};
+```
+
+```go
+func bulbSwitch(n int) int {
+	return int(math.Sqrt(float64(n)))
+}
+```
+
+```ts
+function bulbSwitch(n: number): number {
+    return Math.floor(Math.sqrt(n));
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/0300-0399/0319.Bulb Switcher/Solution.cpp

@@ -0,0 +1,6 @@
+class Solution {
+public:
+    int bulbSwitch(int n) {
+        return (int) sqrt(n);
+    }
+};

solution/0300-0399/0319.Bulb Switcher/Solution.go

@@ -0,0 +1,3 @@
+func bulbSwitch(n int) int {
+	return int(math.Sqrt(float64(n)))
+}

solution/0300-0399/0319.Bulb Switcher/Solution.py

@@ -1,3 +1,3 @@
 class Solution:
     def bulbSwitch(self, n: int) -> int:
-        return int(n ** (1 / 2))
+        return int(sqrt(n))

solution/0300-0399/0319.Bulb Switcher/Solution.ts

@@ -0,0 +1,3 @@
+function bulbSwitch(n: number): number {
+    return Math.floor(Math.sqrt(n));
+}