题目:
方法:
设置两个dict,键值分别为t,与s中出现的字母
value为出现的次数
遍历两个字符串
对比两个dict如果键值对应的value相等则返回True,value不等或键值不同则返回False
class Solution: def isAnagram(self, s, t): """ :type s: str :type t: str :rtype: bool """ if len(s) != len(t):return False dict1 = {} dict2 = {} for i in range(len(s)): if s[i] in dict1:dict1[s[i]] += 1 else:dict1[s[i]] = 1 if t[i] in dict2:dict2[t[i]] += 1 else:dict2[t[i]] = 1 for i in range(len(s)): if s[i] not in dict2:return False if dict1[s[i]] != dict2[s[i]]:return False return True