leetcode之Random Note

几天没做，先做道简单的哈。这题就是将第一个string分解，如果第2个string没有相应的字母，或者字母的数量少于第一个，就说明不匹配。用python的collections包来解决。

from collections import Counter
class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        if ransomNote == '':
            return True
        if magazine == '':
            return False
        note = Counter(ransomNote)
        mag = Counter(magazine)
        for i in note.keys():
            print i
            if None == mag.get(i) or note[i] > mag.get(i):
                return False
        else:
            return True