383. Ransom Note --leetcode

编程练习，多多思考。我的思路是使用字典，通过记录字符对应的值来完成，后来发现 leetcode 上有许多好的思路，很好的利用 python 的优势，学习到了许多。

题目

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

python解法

我觉得这三种其实思路都是一样的，只是在代码编写方面不同，第二种和第三种代码编写方式要好很多，速度也更快，学习到了许多

使用字典

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class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        dic = {}
        for i in magazine:
            if i in dic:
                dic[i] += 1
            else:
                dic[i] = 1
                
        for j in ransomNote:
            if j in dic:
                dic[j] -= 1
                if dic[j] < 0:
                    return False
            else:
                return False
        return True

使用 python 的 set 和 count()

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class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        for i in set(ransomNote):
            if ransomNote.count(i) > magazine.count(i):
                return False
        return True

使用 collections.Counter()

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