题目描述：
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
题目大意：ransom Note构成一个字符串，magazines构成一个字符串，对于ransom Note字符串，能否由magazines中的字符构成，其中magazines中的每个字符只能使用一次。
思路：
1.ransomNote.length > magazines.length 显然，ransomNote不能由magazines构成
2.ransomNote中出现的每一个字符进行统计，统计其出现次数，对magazines中的字符也进行同样的统计。对每一个相同字符，ransomNote中出现的次数小于等于magazines中出现的次数时， ransomNote可由magazines构成
代码：
class Solution(object):
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
if len(ransomNote) > len(magazine):
return False
a = dict()
b = dict()
for i in range(len(ransomNote)):
a[ord(ransomNote[i]) - ord('a')] = 0
for x in range(len(magazine)):
b[ord(magazine[x]) - ord('a')] = 0
for i in range(len(ransomNote)):
a[ord(ransomNote[i]) - ord('a')] +=1
for x in range(len(magazine)):
b[ord(magazine[x]) - ord('a')] +=1
for i in range(len(ransomNote)):
if b.has_key(ord(ransomNote[i]) - ord('a')):
if a[ord(ransomNote[i]) - ord('a')] > b[ord(ransomNote[i]) - ord('a')]:
return False
else:
pass
else:
return False
return True