6/1 每日一題(異位字串)

 Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

 

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true

Example 2:

Input: s = "rat", t = "car"
Output: false

 

Constraints:

• 1 <= s.length, t.length <= 5 * 104
• s and t consist of lowercase English letters.

 

Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

#法一

class Solution:

    def isAnagram(self, s: str, t: str) -> bool:

       

        if len(s) != len(t):

            return False

        s_set=set() #計算元素

       

        for i in s:

            s_set.add(i)

           

        for i in s_set:

            if s.count(i) != t.count(i):

                return False

        return True

#法二

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        
        # if len(s) != len(t):
        #     return False
        # s_set=set() #計算元素
        
        # for i in s:
        #     s_set.add(i)
            
        # for i in s_set:
        #     if s.count(i) != t.count(i):
        #         return False
        # return True
        return Counter(s) == Counter(t)

#法三

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        
        # if len(s) != len(t):
        #     return False
        # s_set=set() #計算元素
        
        # for i in s:
        #     s_set.add(i)
            
        # for i in s_set:
        #     if s.count(i) != t.count(i):
        #         return False
        # return True
        # return Counter(s) == Counter(t)
        a=sorted(s)
        b=sorted(t)
        return a==b