feat: add solutions to lc problems

* No.1021.Remove Outermost Parentheses
* No.2044.Count Number of Maximum Bitwise-OR Subsets

Author: yanglbme

solution/1000-1099/1021.Remove Outermost Parentheses/README.md

@@ -70,7 +70,20 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def removeOuterParentheses(self, s: str) -> str:
+        ans = []
+        cnt = 0
+        for c in s:
+            if c == '(':
+                cnt += 1
+                if cnt > 1:
+                    ans.append(c)
+            else:
+                cnt -= 1
+                if cnt > 0:
+                    ans.append(c)
+        return ''.join(ans)
 ```
 
 ### **Java**
@@ -122,6 +135,29 @@ public:
 };
 ```
 
+### **Go**
+
+```go
+func removeOuterParentheses(s string) string {
+	ans := []rune{}
+	cnt := 0
+	for _, c := range s {
+		if c == '(' {
+			cnt++
+			if cnt > 1 {
+				ans = append(ans, c)
+			}
+		} else {
+			cnt--
+			if cnt > 0 {
+				ans = append(ans, c)
+			}
+		}
+	}
+	return string(ans)
+}
+```
+
 ### **TypeScript**
 
 ```ts

solution/1000-1099/1021.Remove Outermost Parentheses/README_EN.md

@@ -63,7 +63,20 @@ After removing outer parentheses of each part, this is "" + "&quo
 ### **Python3**
 
 ```python
-
+class Solution:
+    def removeOuterParentheses(self, s: str) -> str:
+        ans = []
+        cnt = 0
+        for c in s:
+            if c == '(':
+                cnt += 1
+                if cnt > 1:
+                    ans.append(c)
+            else:
+                cnt -= 1
+                if cnt > 0:
+                    ans.append(c)
+        return ''.join(ans)
 ```
 
 ### **Java**
@@ -113,6 +126,29 @@ public:
 };
 ```
 
+### **Go**
+
+```go
+func removeOuterParentheses(s string) string {
+	ans := []rune{}
+	cnt := 0
+	for _, c := range s {
+		if c == '(' {
+			cnt++
+			if cnt > 1 {
+				ans = append(ans, c)
+			}
+		} else {
+			cnt--
+			if cnt > 0 {
+				ans = append(ans, c)
+			}
+		}
+	}
+	return string(ans)
+}
+```
+
 ### **TypeScript**
 
 ```ts

solution/1000-1099/1021.Remove Outermost Parentheses/Solution.go

@@ -0,0 +1,18 @@
+func removeOuterParentheses(s string) string {
+	ans := []rune{}
+	cnt := 0
+	for _, c := range s {
+		if c == '(' {
+			cnt++
+			if cnt > 1 {
+				ans = append(ans, c)
+			}
+		} else {
+			cnt--
+			if cnt > 0 {
+				ans = append(ans, c)
+			}
+		}
+	}
+	return string(ans)
+}

solution/1000-1099/1021.Remove Outermost Parentheses/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def removeOuterParentheses(self, s: str) -> str:
+        ans = []
+        cnt = 0
+        for c in s:
+            if c == '(':
+                cnt += 1
+                if cnt > 1:
+                    ans.append(c)
+            else:
+                cnt -= 1
+                if cnt > 0:
+                    ans.append(c)
+        return ''.join(ans)

solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/README.md

@@ -58,8 +58,16 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：DFS**
+
 简单 DFS。可以预先算出按位或的最大值 mx，然后 DFS 搜索按位或结果等于 mx 的所有子集数。也可以在 DFS 搜索中逐渐更新 mx 与对应的子集数。
 
+时间复杂度 $O(2^n)$。
+
+**方法二：二进制枚举**
+
+时间复杂度 $O(n*2^n)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -105,6 +113,25 @@ class Solution:
         return ans
 ```
 
+```python
+class Solution:
+    def countMaxOrSubsets(self, nums: List[int]) -> int:
+        n = len(nums)
+        ans = 0
+        mx = 0
+        for mask in range(1 << n):
+            t = 0
+            for i, v in enumerate(nums):
+                if (mask >> i) & 1:
+                    t |= v
+            if mx < t:
+                mx = t
+                ans = 1
+            elif mx == t:
+                ans += 1
+        return ans
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -166,6 +193,31 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int countMaxOrSubsets(int[] nums) {
+        int n = nums.length;
+        int ans = 0;
+        int mx = 0;
+        for (int mask = 1; mask < 1 << n; ++mask) {
+            int t = 0;
+            for (int i = 0; i < n; ++i) {
+                if (((mask >> i) & 1) == 1) {
+                    t |= nums[i];
+                }
+            }
+            if (mx < t) {
+                mx = t;
+                ans = 1;
+            } else if (mx == t) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
+```
+
 ### **TypeScript**
 
 ```ts
@@ -271,6 +323,35 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int countMaxOrSubsets(vector<int>& nums) {
+        int n = nums.size();
+        int ans = 0;
+        int mx = 0;
+        for (int mask = 1; mask < 1 << n; ++mask)
+        {
+            int t = 0;
+            for (int i = 0; i < n; ++i)
+            {
+                if ((mask >> i) & 1)
+                {
+                    t |= nums[i];
+                }
+            }
+            if (mx < t)
+            {
+                mx = t;
+                ans = 1;
+            }
+            else if (mx == t) ++ans;
+        }
+        return ans;
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -297,6 +378,29 @@ func countMaxOrSubsets(nums []int) int {
 }
 ```
 
+```go
+func countMaxOrSubsets(nums []int) int {
+	n := len(nums)
+	ans := 0
+	mx := 0
+	for mask := 1; mask < 1<<n; mask++ {
+		t := 0
+		for i, v := range nums {
+			if ((mask >> i) & 1) == 1 {
+				t |= v
+			}
+		}
+		if mx < t {
+			mx = t
+			ans = 1
+		} else if mx == t {
+			ans++
+		}
+	}
+	return ans
+}
+```
+
 ```go
 func countMaxOrSubsets(nums []int) int {
 	mx, ans := 0, 0