383. Ransom Note（python+cpp）

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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

解释：
数出现的次数即可。
python代码：

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

c++代码：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        set<char> _set(ransomNote.begin(),ransomNote.end());
        for(auto letter:_set)
        {
          if(count(ransomNote.begin(),ransomNote.end(),letter)>count(magazine.begin(),magazine.end(),letter))
                return false; 
        }
        return true;  
    }
};

总结：
注意stl很多函数传入的都是.begin()，.end()。