变位词-全检索-字典

def anagramSolution5(s1,s2):

    list1 = [x for x in range(24)]
    c1 = {key: 0 for key in list1}
    c2 = {key: 0 for key in list1}

    for i in range(len(s1)):
        pos = ord(s1[i]) - ord('a')
        if pos in c1:  # 存在
            c1[pos] += 1  # 叠加
        else:
            c1[pos] = 1  # 不存在，再新建
            c2[pos] = 0  #新建

    for i in range(len(s2)):
        pos = ord(s2[i]) - ord('a')
        if pos in c2:  # 存在
            c2[pos] += 1  # 叠加
        else:
            if pos in c1:
                c1[pos] += 1
            else:
                return False
            c2[pos] = 1  # 不存在，再新建


    for key in c1:
        if c1[key] == c2[key]:
            continue
        else:
            return False

    return True


print("字典",anagramSolution5('中文字符','字符中文'))
print("字典",anagramSolution5('中文强大2e','中文强大字符'))
print("字典",anagramSolution5('中文强大','中文强大##'))
print("字典",anagramSolution5('中文强大##','中文强大##'))