擅长:python、mysql、java
<p>我认为,理想情况下,我们可以使用字典来记录所有字母,并在以后检查</p>
<p>代码:-</p>
<pre><code>def counter(s):
string_list = []
for ch in s.lower():
string_list.append(ch)
string_dict = {}
for c in string_list:
if c not in string_dict:
string_dict[c] = 1
else:
string_dict[c] = string_dict[c] + 1
return string_dict
s1 = "anagram"
s2 = "nagaram"
a = counter(s1)
b = counter(s2)
if a == b:
print("Anagram")
else:
print("Not Anagram")
</code></pre>
<p>使用某种排序函数(如sorted)的计算成本更高</p>