Update 0017.电话号码的字母组合.md

补充python注释规范格式

Author: casnz1601

problems/0017.电话号码的字母组合.md

@@ -282,61 +282,74 @@ class Solution {
 ```
 
 ## Python 
-
-```Python
+**回溯**
+```python3
 class Solution:
-    ans = []
-    s = ''
-    letterMap = {
-        '2': 'abc',
-        '3': 'def',
-        '4': 'ghi',
-        '5': 'jkl',
-        '6': 'mno',
-        '7': 'pqrs',
-        '8': 'tuv',
-        '9': 'wxyz'
-    }
+    def __init__(self):
+        self.answers: List[str] = []
+        self.answer: str = ''
+        self.letter_map = {
+            '2': 'abc',
+            '3': 'def',
+            '4': 'ghi',
+            '5': 'jkl',
+            '6': 'mno',
+            '7': 'pqrs',
+            '8': 'tuv',
+            '9': 'wxyz'
+        }
 
-    def letterCombinations(self, digits):
-        self.ans.clear()
-        if digits == '':
-            return self.ans
+    def letterCombinations(self, digits: str) -> List[str]:
+        self.answers.clear()
+        if not digits: return []
         self.backtracking(digits, 0)
-        return self.ans
-
-    def backtracking(self, digits, index):
-        if index == len(digits):
-            self.ans.append(self.s)
-            return
-        else:
-            letters = self.letterMap[digits[index]]  # 取出数字对应的字符集
-            for letter in letters:
-                self.s = self.s + letter  # 处理
-                self.backtracking(digits, index + 1)
-                self.s = self.s[:-1]      # 回溯
+        return self.answers
+    
+    def backtracking(self, digits: str, index: int) -> None:
+        # 回溯函数没有返回值
+        # Base Case
+        if index == len(digits):    # 当遍历穷尽后的下一层时
+            self.answers.append(self.answer)
+            return 
+        # 单层递归逻辑  
+        letters: str = self.letter_map[digits[index]]
+        for letter in letters:
+            self.answer += letter   # 处理
+            self.backtracking(digits, index + 1)    # 递归至下一层
+            self.answer = self.answer[:-1]  # 回溯
 ```
-
-python3：
-
-```py
+**回溯简化**
+```python3
 class Solution:
+    def __init__(self):
+        self.answers: List[str] = []
+        self.letter_map = {
+            '2': 'abc',
+            '3': 'def',
+            '4': 'ghi',
+            '5': 'jkl',
+            '6': 'mno',
+            '7': 'pqrs',
+            '8': 'tuv',
+            '9': 'wxyz'
+        }
+
     def letterCombinations(self, digits: str) -> List[str]:
-        res = []
-        s = ""
-        letterMap = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
-        if not len(digits): return res
-        def backtrack(digits,index, s):
-            if index == len(digits):
-                return res.append(s)
-            digit = int(digits[index])  #将index指向的数字转为int
-            letters = letterMap[digit]  #取数字对应的字符集
-            for i in range(len(letters)):
-                s += letters[i]
-                backtrack(digits, index+1, s)  #递归，注意index+1，一下层要处理下一个数字
-                s = s[:-1]  #回溯
-        backtrack(digits, 0, s)
-        return res
+        self.answers.clear()
+        if not digits: return []
+        self.backtracking(digits, 0, '')
+        return self.answers
+    
+    def backtracking(self, digits: str, index: int, answer: str) -> None:
+        # 回溯函数没有返回值
+        # Base Case
+        if index == len(digits):    # 当遍历穷尽后的下一层时
+            self.answers.append(answer)
+            return 
+        # 单层递归逻辑  
+        letters: str = self.letter_map[digits[index]]
+        for letter in letters:
+            self.backtracking(digits, index + 1, answer + letter)    # 递归至下一层 + 回溯
 ```