[LeetCode][Python]383. Ransom No

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

思路：

1. 第一反应是依次遍历ransomNote，如果其中的元素都在magazine，则返回True，否则返回False，这个应该可以解决这个问题，不过效率肯定不高。由于元素只能用一次，还要考虑删除的问题。
2. 使用pop和remove同时处理两个列表，对第一个使用pop()，如果在magazine，则删除之。如果元素不在，则返回False，如果在里面，继续删除。
3. 使用collections.Counter()，对于两个Counter的相减，只保留正值的计数。如果每一个value都是正数，则说明magazine可以组成前者。

#!/usr/bin/env python
# -*- coding: UTF-8 -*-
class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        ransomNote = list(ransomNote)
        magazine = list(magazine)
        while ransomNote:
            tem = ransomNote.pop()
            print tem
            if tem not in magazine:
                return False
            else:
                magazine.remove(tem)

        return True

    def canConstruct2(self, ransomNote, magazine):
        import collections
        return not collections.Counter(ransomNote) - collections.Counter(magazine)


    def canConstruct3(self, ransomNote, magazine):
        import collections
        c = collections.Counter(magazine)
        c.subtract(collections.Counter(ransomNote))
        return all(v>=0 for v in c.values())

if __name__ == '__main__':
    sol = Solution()
    s1 = "aa"
    s2 = "aab"
    print sol.canConstruct(s1, s2)
    print sol.canConstruct2(s1, s2)
    print sol.canConstruct3(s1, s2)