383. Ransom Note

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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

简单来说就是判断右边的字符集合是否包含左边的字符集合。直接用数组计数来实现。

代码：

class Solution {public:    bool canConstruct(string ransomNote, string magazine) {    int count[300];    memset(count,0,sizeof(count));    for(auto& x:magazine)    {    count[x]++;    }    for(auto& x:ransomNote)    {    if(count[x]>0)    {    count[x]--;    }    else    {    return false;    }    }    return true;    }};

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