題目描述：

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

題目大意：ransom Note構(gòu)成一個字符串，magazines構(gòu)成一個字符串，對于ransom Note字符串，能否由magazines中的字符構(gòu)成，其中magazines中的每個字符只能使用一次。

思路：

1.ransomNote.length > magazines.length 顯然，ransomNote不能由magazines構(gòu)成

2.ransomNote中出現(xiàn)的每一個字符進行統(tǒng)計，統(tǒng)計其出現(xiàn)次數(shù)，對magazines中的字符也進行同樣的統(tǒng)計。對每一個相同字符，ransomNote中出現(xiàn)的次數(shù)小于等于magazines中出現(xiàn)的次數(shù)時， ransomNote可由magazines構(gòu)成

代碼：

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