6666666666666666666666666

有收获

nan

学习中。

set(string)—返回字符串的独立字符

慢慢来吧

class Rotation(object):

def is_substring(self, s1, s2):
    return s1 in s2

def is_rotation(self, s1, s2):
    if s1 is None or s2 is None:
        return False
    if len(s1) != len(s2):
        return False
    return self.is_substring(s1, s2 + s2)

class Rotation(object):

def is_substring(self, s1, s2):
    return s1 in s2

def is_rotation(self, s1, s2):
    if s1 is None or s2 is None:
        return False
    if len(s1) != len(s2):
        return False
    return self.is_substring(s1, s2 + s2)

class CompressString(object):

def compress(self, string):
    if string is None or not string:
        return string
    result = ''
    prev_char = string[0]
    count = 0
    for char in string:
        if char == prev_char:
            count += 1
        else:
            result += self._calc_partial_result(prev_char, count)
            prev_char = char
            count = 1
    result += self._calc_partial_result(prev_char, count)
    return result if len(result) < len(string) else string

def _calc_partial_result(self, prev_char, count):
    return prev_char + (str(count) if count > 1 else '')

class UniqueChars(object): def has_unique_chars(self, string): if string: return len(set(string)) == len(string) return False

一定要跟你們的答案才能 pass?

!/usr/bin/python3

class UniqueChars(object): def has_unique_chars(self,string): if string == None: return False return len(set(string)) == len(string) """set()函数是创建一个无不重复的元素集"""

非常简单

使用了滑动指针

这个题难在None和“”的判断

非常的简单