[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;
}