【leetcode】925.Long Pressed Name

题目如下：

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:

1. name.length <= 1000
2. typed.length <= 1000
3. The characters of name and typed are lowercase letters.

解题思路：我的方法是把字符串解析成 [字符,该字符连续出现的次数]的格式，例如lleeelee解析的结果是 ['l','2','e','3','l','1','e','2']，最后只要比较name的typed的解析结果即可，比较的方法是字符必须相同，数字必须name <= typed。

代码如下：

class Solution(object):
    def splitStr(self,sval):
        lsplit = []
        lastChar = None
        lastCount = 0
        for i in sval:
            if lastChar == None:
                lastChar = i
                lastCount = 1
            elif lastChar != i:
                lsplit.append(lastChar)
                lsplit.append(str(lastCount))
                lastChar = i
                lastCount = 1
            else:
                lastCount += 1
        lsplit.append(lastChar)
        lsplit.append(str(lastCount))
        return lsplit
    def isLongPressedName(self, name, typed):
        """
        :type name: str
        :type typed: str
        :rtype: bool
        """
        nsplit = self.splitStr(name)
        tsplit = self.splitStr(typed)
        if len(nsplit) != len(tsplit):
            return False
        for v1,v2 in zip(nsplit,tsplit):
            if v1.isdigit() and v2.isdigit():
                if int(v1) > int(v2):
                    return False
            elif v1 != v2:
                return False
        return True