383. Ransom Note

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：
本题题目描述很长，但是总结起来就是一句话：寻找两个字符串中的交叉项（保留重复值），代码思路与前面一篇博客：Intersection
of Two Arrays II惊人相似

代码：

class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
for(int j = 0;j<ransomNote.size();j++)
{
string::iterator itr = find(magazine.begin(),magazine.end(),ransomNote[j]);
if(itr!=magazine.end())
{
magazine.erase(itr);
}
else
return false;

}
return true;

}
};