diff --git a/problems/0746.使用最小花费爬楼梯.md b/problems/0746.使用最小花费爬楼梯.md
index 5931fc8a..abaeb980 100644
--- a/problems/0746.使用最小花费爬楼梯.md
+++ b/problems/0746.使用最小花费爬楼梯.md
@@ -305,5 +305,35 @@ int minCostClimbingStairs(int* cost, int costSize){
     return dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2];
 }
 ```
+
+### Scala
+
+```scala
+object Solution {
+  def minCostClimbingStairs(cost: Array[Int]): Int = {
+    var dp = new Array[Int](cost.length)
+    dp(0) = cost(0)
+    dp(1) = cost(1)
+    for (i <- 2 until cost.length) {
+      dp(i) = math.min(dp(i - 1), dp(i - 2)) + cost(i)
+    }
+    math.min(dp(cost.length - 1), dp(cost.length - 2))
+  }
+}
+```
+
+第二种思路: dp[i] 表示爬到第i-1层所需的最小花费，状态转移方程为: dp[i] = min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2])
+```scala
+object Solution {
+  def minCostClimbingStairs(cost: Array[Int]): Int = {
+    var dp = new Array[Int](cost.length + 1)
+    for (i <- 2 until cost.length + 1) {
+      dp(i) = math.min(dp(i - 1) + cost(i - 1), dp(i - 2) + cost(i - 2))
+    }
+    dp(cost.length)
+  }
+}
+```
+
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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>