Merge pull request #1483 from wzqwtt/greedy08

添加（0714.买卖股票的最佳时机含手续费、0968.监控二叉树） Scala版本

problems/0714.买卖股票的最佳时机含手续费.md

@@ -153,7 +153,7 @@ public:
 
 ## 其他语言版本
 
-Java：
+### Java
 ```java
 // 贪心思路
 class Solution {
@@ -198,7 +198,7 @@ class Solution { // 动态规划
 
 
 
-Python：
+### Python
 
 ```python
 class Solution: # 贪心思路
@@ -216,7 +216,7 @@ class Solution: # 贪心思路
         return result
 ```
 
-Go：
+### Go
 ```golang
 func maxProfit(prices []int, fee int) int {
     var minBuy int = prices[0] //第一天买入
@@ -241,7 +241,7 @@ func maxProfit(prices []int, fee int) int {
     return res
 }
 ```
-Javascript:
+### Javascript
 ```Javascript
 // 贪心思路
 var maxProfit = function(prices, fee) {
@@ -293,7 +293,7 @@ var maxProfit = function(prices, fee) {
 };
 ```
 
-TypeScript：
+### TypeScript
 
 > 贪心
 
@@ -335,8 +335,28 @@ function maxProfit(prices: number[], fee: number): number {
 };
 ```
 
+### Scala
 
+贪心思路:
 
+```scala
+object Solution {
+  def maxProfit(prices: Array[Int], fee: Int): Int = {
+    var result = 0
+    var minPrice = prices(0)
+    for (i <- 1 until prices.length) {
+      if (prices(i) < minPrice) {
+        minPrice = prices(i) // 比当前最小值还小
+      }
+      if (prices(i) > minPrice + fee) {
+        result += prices(i) - minPrice - fee
+        minPrice = prices(i) - fee
+      }
+    }
+    result
+  }
+}
+```
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0968.监控二叉树.md

@@ -544,5 +544,40 @@ int minCameraCover(struct TreeNode* root){
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  def minCameraCover(root: TreeNode): Int = {
+    var result = 0
+    def traversal(cur: TreeNode): Int = {
+      // 空节点，该节点有覆盖
+      if (cur == null) return 2
+      var left = traversal(cur.left)
+      var right = traversal(cur.right)
+      // 情况1，左右节点都有覆盖
+      if (left == 2 && right == 2) {
+        return 0
+      }
+      // 情况2
+      if (left == 0 || right == 0) {
+        result += 1
+        return 1
+      }
+      // 情况3
+      if (left == 1 || right == 1) {
+        return 2
+      }
+      -1
+    }
+
+    if (traversal(root) == 0) {
+      result += 1
+    }
+    result
+  }
+}
+```
+
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>