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") -> falsecanConstruct("aa", "ab") -> falsecanConstruct("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;              }};