long pressed name easy

Author: weiy

String/LongPressedName.py

@@ -0,0 +1,67 @@
+"""
+Your friend is typing his name into a keyboard.  Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.
+
+You examine the typed characters of the keyboard.  Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
+
+ 
+
+Example 1:
+
+Input: name = "alex", typed = "aaleex"
+Output: true
+Explanation: 'a' and 'e' in 'alex' were long pressed.
+Example 2:
+
+Input: name = "saeed", typed = "ssaaedd"
+Output: false
+Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
+Example 3:
+
+Input: name = "leelee", typed = "lleeelee"
+Output: true
+Example 4:
+
+Input: name = "laiden", typed = "laiden"
+Output: true
+Explanation: It's not necessary to long press any character.
+ 
+
+Note:
+
+name.length <= 1000
+typed.length <= 1000
+The characters of name and typed are lowercase letters.
+
+判断 typed 里 是否存在 name，保证顺序的。
+
+思路：
+typed 和 name 各一个指针，命中了就一起走，没命中就typed走。最后判断指针是否指向了name尾。
+
+测试地址：
+https://leetcode.com/contest/weekly-contest-107/problems/long-pressed-name/
+
+"""
+class Solution(object):
+    def isLongPressedName(self, name, typed):
+        """
+        :type name: str
+        :type typed: str
+        :rtype: bool
+        """
+        x = 0
+        y = 0
+        
+        x_length = len(name)
+        y_length = len(typed)
+        
+        while x < x_length and y < y_length:
+            
+            if typed[y] == name[x]:
+                y += 1
+                x += 1
+            else:
+                y += 1
+        
+        if x == x_length:
+            return True
+        return False