算法面试 字符串全排列

def perm(s=''):
    if len(s) <= 1:
        return [s]
    sl = []
    for i in range(len(s)):
        for j in perm(s[0:i] + s[i + 1:]):
            sl.append(s[i] + j)
    return sl


def main():
    # 可能包含重复的串
    perm_nums = perm('abb')
    # 对结果去重
    no_repeat_nums = list(set(perm_nums))
    print('perm_nums', len(perm_nums), perm_nums)
    print('no_repeat_nums', len(no_repeat_nums), no_repeat_nums)
    pass


if __name__ == '__main__':
    main()
原文地址:https://www.cnblogs.com/cupleo/p/10870513.html