<LeetCode OJ> 383. Ransom Note

版权声明：本文为EbowTang原创文章，后续可能继续更新本文。如果转载，请务必复制本文末尾的信息！ https://blog.csdn.net/EbowTang/article/details/52187660

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

分析：DONE

判断给定字符串Ransom 能否由另一字符串magazine中字符中的某些字符组合生成。

显然，统计字符出现次数即可！

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> charcnt(26,0);
        //统计magazine中每个字符出现次数
        for(int i=0;i<magazine.size();i++)
            charcnt[magazine[i]-'a']++;
        //统计ransomNote中每个字符出现次数    
        for(int i=0;i<ransomNote.size();i++)
            charcnt[ransomNote[i]-'a']--;
        //检查是否ransomNote中的数量是否都小于magazine中的！
        for(int i=0;i<ransomNote.size();i++)
            if(charcnt[ransomNote[i]-'a'] < 0)
                return false;
        return true;
    }
};

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/52187660

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895