zoukankan      html  css  js  c++  java
  • 125-242. 有效的字母异位词

    给定两个字符串 s 和 t ,编写一个函数来判断 t 是否是 s 的字母异位词。 (照旧第一个我写的,其他的我抄的.)
    class Solution(object):
        def isAnagram1(self, s, t):
            """
            :type s: str
            :type t: str
            :rtype: bool
            """
            set_s = set(s)
            set_t = set(t)
            if len(s) != len(t) or set_s != set_t:
                return False
            s = sorted(s)
            t = sorted(t)
            return s == t
    
        def isAnagram2(self, s, t):
            """ 我可能是疯了,我一直感觉count很内存所以没用,我想多了
            :type s: str
            :type t: str
            :rtype: bool
            """
            if len(s) != len(t):
                return False
            a = set(s)
            for i in a:
                if s.count(i) != t.count(i):
                    return False
            return True
    
        def isAnagram(self, s, t):
            """  实验才是检验真理的唯一标椎
            :type s: str
            :type t: str
            :rtype: bool
            """
            s_dic, t_dic = {}, {}
            for i in s:
                s_dic[i] = s_dic.get(i, 0) + 1
    
            for i in t:
                t_dic[i] = t_dic.get(i, 0) + 1
    
            return s_dic == t_dic
    
    
    if __name__ == '__main__':
        s1 = Solution()
        # s = "rat"; t = "car"
        # s = "anagram"; t = "nagaram"
        # s = "aa"; t = "bb"
        s = "aabbbb"; t="aaaabb"
        print(s1.isAnagram(s, t))
    
  • 相关阅读:
    解决一切日期问题的日期类
    汉诺塔类型问题解析
    窗口滑动
    大暑假集训总结(反思)
    找硬币题解
    Fiolki题解
    大逃亡题解
    Luogu6080 [USACO05DEC]Cow Patterns G
    Luogu3193 HNOI2008 GT考试
    Codeforces1355F Guess Divisors Count
  • 原文地址:https://www.cnblogs.com/liuzhanghao/p/14228591.html
Copyright © 2011-2022 走看看