LeetCode242：Valid Anagram

Given two strings s and t , write a function to determine if t is an anagram of s.

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true

Example 2:

Input: s = "rat", t = "car"
Output: false

Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

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LeetCode：链接

直观的思路是，统计两个字符串中不同字符出现的次数，只有当两者出现的字符相同且出现的次数相同，那么它们是“anagram”。另外，两个长度不同的字符串一定不满足要求，可以辅助判断。 

通用的方法应该用字典处理，对于unicode也可以处理。分别对s和t生成两个字典，判断字典是否相同即可。

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(s) != len(t):
            return False

        def count(s):
            dict = {}
            for i in range(len(s)):
                if s[i] not in dict:
                    dict[s[i]] = 1
                else:
                    dict[s[i]] += 1
            return dict

        dict_s = count(s)
        dict_t = count(t)
        return dict_s == dict_t

更简单的写法，Python 字典(Dictionary) get() 函数返回指定键的值，如果值不在字典中返回默认值：

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(s) != len(t):
            return False

        alpha = {}
        beta = {}
        for c in s:
            alpha[c] = alpha.get(c, 0) + 1
        for c in t:
            beta[c] = beta.get(c, 0) + 1
        return alpha == beta