0841.钥匙和房间.md, 增加 Golang, Java, Python 实现

problems/0841.钥匙和房间.md

@@ -121,10 +121,109 @@ public:
 
 Java：
 
+```java
+class Solution {
+     private void dfs(int key, List<List<Integer>>  rooms, List<Boolean> visited) {
+        if (visited.get(key)) {
+            return;
+        }
+
+        visited.set(key, true);
+        for (int k : rooms.get(key)) {
+            // 深度优先搜索遍历
+            dfs(k, rooms, visited);
+        }
+    }
+
+
+    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
+        List<Boolean> visited = new ArrayList<Boolean>(){{
+            for(int i = 0 ; i < rooms.size(); i++){
+                add(false);
+            }
+        }};
+
+        dfs(0, rooms, visited);
+
+        //检查是否都访问到了
+        for (boolean flag : visited) {
+            if (!flag) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
+
+
+
 Python：
 
+python3
+
+```python
+
+class Solution:
+
+    def dfs(self, key: int, rooms: List[List[int]]  , visited : List[bool] ) :
+        if visited[key] :
+            return
+
+        visited[key] = True
+        keys = rooms[key]
+        for i in range(len(keys)) :
+            # 深度优先搜索遍历
+            self.dfs(keys[i], rooms, visited)
+
+    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
+        visited = [False for i in range(len(rooms))]
+
+        self.dfs(0, rooms, visited)
+
+        # 检查是否都访问到了
+        for i in range(len(visited)):
+            if not visited[i] :
+                return False
+        return True
+
+```
+
+
 Go：
 
+```go
+
+func dfs(key int, rooms [][]int, visited []bool ) {
+	if visited[key] {
+		return;
+	}
+
+	visited[key] = true
+	keys := rooms[key]
+	for _ , key := range keys {
+		// 深度优先搜索遍历
+		dfs(key, rooms, visited);
+	}
+}
+
+func canVisitAllRooms(rooms [][]int) bool {
+
+	visited := make([]bool, len(rooms));
+
+	dfs(0, rooms, visited);
+
+	//检查是否都访问到了
+	for i := 0; i < len(visited); i++ {
+		if !visited[i] {
+			return false;
+		}
+	}
+	return true;
+}
+```
+
 JavaScript：
 
 -----------------------

problems/0844.比较含退格的字符串.md

@@ -188,8 +188,49 @@ class Solution {
 
 Python：
 
+python3
+
+```python
+class Solution:
+
+    def get_string(self, s: str) -> str :
+        bz = []
+        for i in range(len(s)) :
+            c = s[i]
+            if c != '#' :
+                bz.append(c) # 模拟入栈
+            elif len(bz) > 0: # 栈非空才能弹栈
+                bz.pop() # 模拟弹栈
+        return str(bz)
+
+    def backspaceCompare(self, s: str, t: str) -> bool:
+        return self.get_string(s) == self.get_string(t)
+        pass
+```
+
+
 Go：
 
+```go
+
+func getString(s string) string {
+	bz := []rune{}
+	for _, c := range s {
+		if c != '#' {
+			bz = append(bz, c); // 模拟入栈
+		} else if len(bz) > 0 { // 栈非空才能弹栈
+			bz = bz[:len(bz)-1] // 模拟弹栈
+		}
+	}
+	return string(bz)
+}
+
+func backspaceCompare(s string, t string) bool {
+	return getString(s) == getString(t)
+}
+
+```
+
 JavaScript：
 
 -----------------------

problems/0922.按奇偶排序数组II.md

@@ -11,6 +11,8 @@
 
 # 922. 按奇偶排序数组II
 
+[力扣题目链接](https://leetcode-cn.com/problems/sort-array-by-parity-ii/)
+
 给定一个非负整数数组 A， A 中一半整数是奇数，一半整数是偶数。
 
 对数组进行排序，以便当 A[i] 为奇数时，i 也是奇数；当 A[i] 为偶数时， i 也是偶数。
@@ -147,9 +149,9 @@ class Solution {
 }
 ```
 
-## Python
+## Python3
 
-```python3
+```python
 #方法2
 class Solution:
     def sortArrayByParityII(self, nums: List[int]) -> List[int]:
@@ -180,6 +182,28 @@ class Solution:
 ## Go
 
 ```go
+
+// 方法一
+func sortArrayByParityII(nums []int) []int {
+	// 分别存放 nums 中的奇数、偶数
+	even, odd := []int{}, []int{}
+	for i := 0; i < len(nums); i++ {
+		if (nums[i] % 2 == 0) {
+			even = append(even, nums[i])
+		} else {
+			odd = append(odd, nums[i])
+		}
+	}
+
+	// 把奇偶数组重新存回 nums
+	result := make([]int, len(nums))
+	index := 0
+	for i := 0; i < len(even); i++ {
+		result[index] = even[i]; index++;
+		result[index] = odd[i]; index++;
+	}
+	return result;
+}
 ```
 
 ## JavaScript

problems/0925.长按键入.md

@@ -134,7 +134,7 @@ Python：
 class Solution:
     def isLongPressedName(self, name: str, typed: str) -> bool:
         i, j = 0, 0
-        m, n = len(name) , len(typed) 
+        m, n = len(name) , len(typed)
         while i< m and j < n:
             if name[i] == typed[j]: # 相同时向后匹配
                 i += 1
@@ -155,8 +155,32 @@ class Solution:
             else: return False
         return True
 ```
+
 Go：
 
+```go
+
+func isLongPressedName(name string, typed string) bool {
+	if(name[0] != typed[0] || len(name) > len(typed)) {
+		return false;
+	}
+
+	idx := 0 // name的索引
+	var last byte  // 上个匹配字符
+	for i := 0; i < len(typed); i++ {
+		if idx < len(name) && name[idx] == typed[i] {
+			last = name[idx]
+			idx++
+		} else if last == typed[i] {
+			continue
+		} else  {
+			return false
+		}
+	}
+	return idx == len(name)
+}
+```
+
 JavaScript：
 
 -----------------------

problems/0941.有效的山脉数组.md

@@ -7,7 +7,7 @@
 </p>
 <p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
 
-# 941.有效的山脉数组 
+# 941.有效的山脉数组
 
 [力扣题目链接](https://leetcode-cn.com/problems/valid-mountain-array/)
 
@@ -103,14 +103,52 @@ class Solution {
 }
 ```
 
-## Python
+## Python3
 
 ```python
+class Solution:
+    def validMountainArray(self, arr: List[int]) -> bool:
+        if len(arr) < 3 :
+            return False
+
+        i = 1
+        flagIncrease = False # 上升
+        flagDecrease = False # 下降
+
+        while i < len(arr) and arr[i-1] < arr[i]:
+            flagIncrease = True
+            i += 1
+
+        while i < len(arr) and arr[i-1] > arr[i]:
+            flagDecrease = True
+            i += 1
+
+        return i == len(arr) and flagIncrease and flagDecrease
+
 ```
 
 ## Go
 
 ```go
+func validMountainArray(arr []int) bool {
+	if len(arr) < 3 {
+		return false
+	}
+
+	i := 1
+	flagIncrease := false // 上升
+	flagDecrease := false // 下降
+
+	for ; i < len(arr) && arr[i-1] < arr[i]; i++ {
+		flagIncrease = true;
+	}
+
+	for ; i < len(arr) && arr[i-1] > arr[i]; i++ {
+		flagDecrease = true;
+	}
+
+	return i == len(arr) && flagIncrease && flagDecrease;
+}
 ```
 
 ## JavaScript