diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md
index 2b5cb978..e357a33b 100644
--- a/problems/0017.电话号码的字母组合.md
+++ b/problems/0017.电话号码的字母组合.md
@@ -557,6 +557,37 @@ func letterCombinations(_ digits: String) -> [String] {
 }
 ```
 
+## Scala:
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def letterCombinations(digits: String): List[String] = {
+    var result = mutable.ListBuffer[String]()
+    if(digits == "") return result.toList // 如果参数为空，返回空结果集的List形式
+    var path = mutable.ListBuffer[Char]()
+    // 数字和字符的映射关系
+    val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz")
+
+    def backtracking(index: Int): Unit = {
+      if (index == digits.size) {
+        result.append(path.mkString)  // mkString语法：将数组类型直接转换为字符串
+        return
+      }
+      var digit = digits(index) - '0' // 这里使用toInt会报错！必须 -'0'
+      for (i <- 0 until map(digit).size) {
+        path.append(map(digit)(i))
+        backtracking(index + 1)
+        path = path.take(path.size - 1)
+      }
+    }
+    
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index 0ace0fd5..2c1bf717 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -502,5 +502,35 @@ func combinationSum3(_ count: Int, _ targetSum: Int) -> [[Int]] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def combinationSum3(k: Int, n: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(k: Int, n: Int, sum: Int, startIndex: Int): Unit = {
+      if (sum > n) return // 剪枝，如果sum>目标和，就返回
+      if (sum == n && path.size == k) {
+        result.append(path.toList)
+        return
+      }
+      // 剪枝
+      for (i <- startIndex to (9 - (k - path.size) + 1)) {
+        path.append(i)
+        backtracking(k, n, sum + i, i + 1) 
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(k, n, 0, 1) // 调用递归方法
+    result.toList // 最终返回结果集的List形式
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>