383. Ransom Note 难度：easy

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

思路：

列出了magazine的字母表，然后算出了出现个数，然后遍历ransomNote，保证有足够的字母可用。

程序：

class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
vector<int> count(26,0);
for(int i = 0;i < magazine.size();i++)
count[magazine[i] - 'a']++;

for(int i = 0;i < ransomNote.size();i++)
{
if(!(count[ransomNote[i] - 'a']--))
return false;
}
return true;
}
};