# 解题思路：字典解决其对应关系 20190302 找工作期间

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        #字母异位词指字母相同，但排列不同的字符串


        # dic1=collections.Counter(s)
        # dic2=collections.Counter(t)

        dic1 = dict()
        dic2 = dict()

        for i in range(len(s)):
            if s[i] not in dic1:
                dic1[s[i]] = 1
            else:
                dic1[s[i]] += 1

        for j in range(len(t)):
            if t[j] not in dic2:
                dic2[t[j]] = 1
            else:
                dic2[t[j]] += 1

        if len(dic1)!=len(dic2):#此种情况直接返回False
            return False
        for key,value in dic2.items():
            if key in dic1 and value==dic1[key]:#key在dic1中并且值相等
                continue
            else:
                return False
        return True