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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
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