[leetcode]: 383. Ransom Note

1.题目描述

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.
翻译：给两个字符串note，magazine。求：是否能通过magazine中的字符重新组合成note。（magazine中每个字符只能用一次，假设只有小写字母）
例：
canConstruct(“a”, “b”) -> false
canConstruct(“aa”, “ab”) -> false
canConstruct(“aa”, “aab”) -> true

2.分析

统计note,magazine中各个字符出现的个数。对于需要的字符,magazine出现的次数不小于在note中出现的次数即可。
次数统计可以手动统计，也可以使用容器。

3.代码

c++

bool canConstruct(string ransomNote, string magazine) {
    int* hashTable = new int[26]();
    for (char ch : magazine)
        ++hashTable[ch - 'a'];
    for (char ch : ransomNote) {
        --hashTable[ch - 'a'];
        if (hashTable[ch - 'a'] < 0)
            return false;
    }
    return true;
}

或者

bool canConstruct(string ransomNote, string magazine) {
    unordered_multiset<char> note(ransomNote.begin(), ransomNote.end());
    unordered_multiset<char> magz(magazine.begin(), magazine.end());
    for (char c : ransomNote)
        if (note.count(c) > magz.count(c))
            return false;
    return true;
}