383. Ransom Note (Easy)

Posted Yancea --- TO BE A DOER !

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了383. Ransom Note (Easy)相关的知识,希望对你有一定的参考价值。

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

题意:判断magazine中的字符是否可以组成ransom note所需要的那些字符;
思路:
1.利用Python的collections.Counter()统计字符个数,然后做差即可;
2.当ransomCnt大于magazineCnt时,返回false;
class Solution():
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        ransomCnt = collections.Counter(ransomNote)
        magazineCnt = collections.Counter(magazine)
        return not ransomCnt - magazineCnt

以上是关于383. Ransom Note (Easy)的主要内容,如果未能解决你的问题,请参考以下文章

LeetCode 383. Ransom Note

LeetCode(383)Ransom Note

LeetCode 383 Ransom Note

LeetCode 383 Ransom Note

LeetCode:Ransom Note_383

LeetCode_383. Ransom Note