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
非常簡單的一道題，就是用哈希Map統計字符的個數，參見代碼如下：
class Solution { public: bool canConstruct(string ransomNote, string magazine) { unordered_map<char, int> m; for (char c : magazine) ++m[c]; for (char c : ransomNote) { if (--m[c] < 0) return false; } return true; } };
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