LeetCode 925. Long Pressed Name Python 使用re

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.

使用re时间会慢点

import re
class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        tsz=len(typed)
        if len(name) == 0 and not tsz== 0:
            return False
        elif name==typed:
            return True
        str = name[0]
        idx=0
        
        for i in range(1,len(name)):       
            if name[i-1]==name[i]:
                str += name[i]
            else:
                match=re.match(str+'+',typed[idx:])
                if not match:
                    return False
                str=name[i]
                idx+=match.end()
                if idx>tsz-1:
                    return False
        return True
           

不使用re

class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        nsz,tsz=len(name),len(typed)
        if name==typed:
            return True
        elif nsz==0 and not tsz==0:
            return False 
        i = 0
        n=''
        for j in range(tsz):
            if i<nsz and name[i]==typed[j]:
                    i+=1
                    n=name[i-1]
            elif n==typed[j]:
                continue
            else:
                return False
        return i==nsz