383. Ransom Note (Easy)

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

题意：判断magazine中的字符是否可以组成ransom note所需要的那些字符；
思路：
1.利用Python的collections.Counter()统计字符个数，然后做差即可；
2.当ransomCnt大于magazineCnt时，返回false；

class Solution():
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        ransomCnt = collections.Counter(ransomNote)
        magazineCnt = collections.Counter(magazine)
        return not ransomCnt - magazineCnt

383. Ransom Note (Easy)

原文：http://www.cnblogs.com/yancea/p/7508941.html