leetcode 242. 有效的字母异位词

思路就是使用哈希标表，将元素存入表中，对比元素的个数。

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        s1 = {}
        t1 = {}
        if len(s) != len(t):
            return False
        
        for i in s:
            if i in s1:
                s1[i] += 1
            else:
                s1[i] = 0
        for i in t:
            if i in t1:
                t1[i] += 1
            else:
                t1[i] = 0
	   for i in s1.keys():
	        if i in t1.keys():
	            if t1[i] != s1[i]:
	                return False
	        else:
	            return False
	
	    return True

看一种简洁的代码：

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        s1 = {}
        t1 = {}
        for i in s:
            s1[i] = s1.get(i, 0) + 1
        for i in t:
            t1[i] = t1.get(i, 0) + 1
        return s1 == t1

还有更简洁的代码

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        return sorted(s) == sorted(t)