class AnagramDetection:
    # 先对两个字符串进行list化
    # 对字符串对应的两个list进行排序
    # 依次比较字符是否匹配
    def anagramSolution1(self, s1, s2):
        alist1 = list(s1)
        alist2 = list(s2)

        alist1.sort()
        alist2.sort()

        pos = 0
        matches = True

        while pos < len(s1) and matches:
            if alist1[pos] == alist2[pos]:
                pos = pos + 1
            else:
                matches = False

        return matches

    # 首先生成两个26个字母的list
    # 计算每个字母出现的次数并存入到相应的list中
    # 比较两个list是否相同
    def anagramSolution2(self, s1, s2):
        c1 = [0] * 26
        c2 = [0] * 26

        for i in range(len(s1)):
            pos = ord(s1[i]) - ord('a')
            c1[pos] = c1[pos] + 1

        for i in range(len(s2)):
            pos = ord(s2[i]) - ord('a')
            c2[pos] = c2[pos] + 1

        j = 0
        stillOK = True
        while j < 26 and stillOK:
            if c1[j] == c2[j]:
                j = j + 1
            else:
                stillOK = False

        return stillOK

    # 首先将两个字符串list化
    # 将两个list中的字符生成两个set
    # 比较两个set, 如果不相等直接返回false
    # 如果两个set相等, 比较每个set中字符在相应list中的个数, 个数不同返回false
    def anagramSolution3(self, s1, s2):
        alist1 = list(s1)
        alist2 = list(s2)

        aset1 = set(alist1)
        aset2 = set(alist2)

        if aset1 != aset2:
            return False
        else:
            for ch in aset1:
                if alist1.count(ch) != alist2.count(ch):
                    return False
            return True

s1 = 'abcde'
s2 = 'acbde'
test = AnagramDetection()
print(test.anagramSolution1(s1, s2))
print(test.anagramSolution2(s1, s2))
print(test.anagramSolution3(s1, s2))