比较两个字符串是否有相同的字母且相同字母个数相等

问题
若给两组字符串如：
s1 = ‘Public Relations’
s2 = ‘Crap Built on Lies’
如果忽略大小写与空格，s1可以通过打乱顺序得到s2，则返回True，否则返回False

s1 = 'Public Relations'
s2 = 'Crap Built on Lies'

import re
#变为小写字母
s1 = s1.lower()
s2 = s2.lower()    
#正则表达式，去除空格
s1 = re.sub(r"\s+", "", s1, flags=re.UNICODE)  #\s 匹配任意的空白符； + 重复1次或多次
s2 = re.sub(r"\s+", "", s2, flags=re.UNICODE)              

正则表达式参考 正则表达式

#方法一：使用sorted()函数
def anagram1(s1,s2):
    s1 = s1.replace(' ','').lower()
    s2 = s2.replace(' ','').lower()
    return sorted(s1) == sorted(s2)

#方法二
def anagram2(s1,s2):
    s1 = s1.replace(' ','').lower()
    s2 = s2.replace(' ','').lower()
    
    if len(s1) != len(s2):
        return False
    
    count = {}
    
    # fill dictionary with first string
    for letter in s1:
        if letter in count:
            count[letter] += 1
        else:
            count[letter] = 1
            
    for letter in s2:
        if letter in count:
            count[letter] -= 1
        else:
            count[letter] = 1
            
    for k in count:
        if count[k] != 0:
            return False
    
    return True