383. Ransom Note--哈希的方法

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

方法一、采用c++的map来实现，代码：

bool canConstruct(string ransomNote, string magazine) {
//采用c++的map来实现
map<char,int> mapNum;

for(int i = 0; i < magazine.length(); i++){
mapNum[magazine[i]]++;
}

for(int i = 0; i < ransomNote.length(); i++){
if(--mapNum[ransomNote[i]] < 0)
return false;
}
return true;
}

方法二、采用数组来实现，代码：

bool canConstruct(string ransomNote, string magazine) {
vector<int> vec(26, 0);
for (int i = 0; i < magazine.size(); ++i)
++vec[magazine[i] - 'a'];
for (int j = 0; j < ransomNote.size(); ++j)
if (--vec[ransomNote[j] - 'a'] < 0)
return false;
return true;
}