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 0f99c879..58508e08 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
@@ -39,7 +39,7 @@
 
 可以使用map或者定义一个二位数组，例如：string letterMap[10]，来做映射，我这里定义一个二维数组，代码如下：
 
-```
+```cpp
 const string letterMap[10] = {
     "", // 0
     "", // 1
@@ -79,7 +79,7 @@ const string letterMap[10] = {
 
 代码如下：
 
-```
+```cpp
 vector<string> result;
 string s;
 void backtracking(const string& digits, int index)
@@ -95,7 +95,7 @@ void backtracking(const string& digits, int index)
 
 代码如下：
 
-```
+```cpp
 if (index == digits.size()) {
     result.push_back(s);
     return;
@@ -281,7 +281,7 @@ class Solution {
 
 ## Python 
 **回溯**
-```python3
+```python
 class Solution:
     def __init__(self):
         self.answers: List[str] = []
@@ -317,7 +317,7 @@ class Solution:
             self.answer = self.answer[:-1]  # 回溯
 ```
 **回溯简化**
-```python3
+```python
 class Solution:
     def __init__(self):
         self.answers: List[str] = []
@@ -420,7 +420,8 @@ var letterCombinations = function(digits) {
 };
 ```
 
-C:
+## C
+
 ```c
 char* path;
 int pathTop;
@@ -481,6 +482,47 @@ char ** letterCombinations(char * digits, int* returnSize){
 }
 ```
 
+## Swift
+
+```swift
+func letterCombinations(_ digits: String) -> [String] {
+    // 按键与字母串映射
+    let letterMap = [
+        "",
+        "", "abc", "def",
+        "ghi", "jkl", "mno",
+        "pqrs", "tuv", "wxyz"
+    ]
+    // 把输入的按键字符串转成Int数组
+    let baseCode = ("0" as Character).asciiValue!
+    let digits = digits.map { c in
+        guard let code = c.asciiValue else { return -1 }
+        return Int(code - baseCode)
+    }.filter { $0 >= 0 && $0 <= 9 }
+    guard !digits.isEmpty else { return [] }
+
+    var result = [String]()
+    var s = ""
+    func backtracking(digits: [Int], index: Int) {
+        // 结束条件：收集结果
+        if index == digits.count {
+            result.append(s)
+            return
+        }
+
+        // 遍历当前按键对应的字母串
+        let letters = letterMap[digits[index]]
+        for letter in letters {
+            s.append(letter) // 处理
+            backtracking(digits: digits, index: index + 1) // 递归，记得+1
+            s.removeLast() // 回溯
+        }
+    }
+    backtracking(digits: digits, index: 0)
+    return 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>