优化0925.长按键入python版本

problems/0925.长按键入.md

@@ -129,29 +129,21 @@ class Solution {
 ```
 ### Python
 ```python
-class Solution:
-    def isLongPressedName(self, name: str, typed: str) -> bool:
-        i, j = 0, 0
-        m, n = len(name) , len(typed)
-        while i< m and j < n:
-            if name[i] == typed[j]: # 相同时向后匹配
-                i += 1
-                j += 1
-            else: # 不相同
-                if j == 0: return False # 如果第一位不相同，直接返回false
-                # 判断边界为n-1,若为n会越界,例如name:"kikcxmvzi" typed:"kiikcxxmmvvzzz"
-                while j < n - 1  and typed[j] == typed[j-1]: j += 1
-                if name[i] == typed[j]:
-                    i += 1
-                    j += 1
-                else: return False
-        # 说明name没有匹配完
-        if i < m: return False
-        # 说明type没有匹配完
-        while j < n:
-            if typed[j] == typed[j-1]: j += 1
-            else: return False
-        return True
+        i = j = 0
+        while(i<len(name) and j<len(typed)):
+        # If the current letter matches, move as far as possible
+            if typed[j]==name[i]:
+                while j+1<len(typed) and typed[j]==typed[j+1]:
+                    j+=1
+                    # special case when there are consecutive repeating letters
+                    if i+1<len(name) and name[i]==name[i+1]:
+                        i+=1
+                else:
+                    j+=1
+                    i+=1
+            else:
+                return False
+        return i == len(name) and j==len(typed)
 ```
 
 ### Go