diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md
index 41940487..11ca2d03 100644
--- a/problems/0093.复原IP地址.md
+++ b/problems/0093.复原IP地址.md
@@ -659,6 +659,48 @@ func restoreIpAddresses(_ s: String) -> [String] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def restoreIpAddresses(s: String): List[String] = {
+    var result = mutable.ListBuffer[String]()
+    if (s.size < 4 || s.length > 12) return result.toList
+    var path = mutable.ListBuffer[String]()
+
+    // 判断IP中的一个字段是否为正确的
+    def isIP(sub: String): Boolean = {
+      if (sub.size > 1 && sub(0) == '0') return false
+      if (sub.toInt > 255) return false
+      true
+    }
+
+    def backtracking(startIndex: Int): Unit = {
+      if (startIndex >= s.size) {
+        if (path.size == 4) {
+          result.append(path.mkString(".")) // mkString方法可以把集合里的数据以指定字符串拼接
+          return
+        }
+        return
+      }
+      // subString
+      for (i <- startIndex until startIndex + 3 if i < s.size) {
+        var subString = s.substring(startIndex, i + 1)
+        if (isIP(subString)) { // 如果合法则进行下一轮
+          path.append(subString)
+          backtracking(i + 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/0131.分割回文串.md b/problems/0131.分割回文串.md
index a371864d..64d45853 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -676,5 +676,50 @@ impl Solution {
     }
 }
 ```
+
+
+## Scala
+
+```scala
+object Solution {
+
+  import scala.collection.mutable
+
+  def partition(s: String): List[List[String]] = {
+    var result = mutable.ListBuffer[List[String]]()
+    var path = mutable.ListBuffer[String]()
+
+    // 判断字符串是否回文
+    def isPalindrome(start: Int, end: Int): Boolean = {
+      var (left, right) = (start, end)
+      while (left < right) {
+        if (s(left) != s(right)) return false
+        left += 1
+        right -= 1
+      }
+      true
+    }
+
+    // 回溯算法
+    def backtracking(startIndex: Int): Unit = {
+      if (startIndex >= s.size) {
+        result.append(path.toList)
+        return
+      }
+      // 添加循环守卫，如果当前分割是回文子串则进入回溯
+      for (i <- startIndex until s.size if isPalindrome(startIndex, i)) {
+        path.append(s.substring(startIndex, i + 1))
+        backtracking(i + 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>