383. Ransom Note

problem

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

用报纸拼字符串的问题，后面的参数为报纸，前面的为待拼字符串

solution

• 使用collections.Counter

from collections import Counter
class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        if ransomNote:
            target = Counter(ransomNote+'a')
            resource = Counter(magazine+'a')
            return all([True if resource[i]- target[i] >= 0 else False for i in target])
        else:
            return True

discuss里有一个更精简的写法.同样的问题是效率不高（后20%）

反过来减，如果不能完全包含则不为空，反之为空
利用not 把返回值空/非空，转换为True/False

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

discuss

很好理解，效率高一些（可能是in的判断减少了很多无效的比较）

magazine会不断变化剔除用过的元素个体

def canConstruct(self, ransomNote, magazine):
    """
    :type ransomNote: str
    :type magazine: str
    :rtype: bool
    """
    for element in ransomNote:
        if element not in magazine:
            return False
        else:
            i = magazine.find(element)
            magazine = magazine[:i] + magazine[i+1:]
    return True