LeetCode 383. Ransom Note

问题描述：

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：大意就是给两个字符串，判断第一个字符串能否由第二个字串中字母组成，第二个字符串中的每个字符只能使用1次

个人用的hash表解决，0-256的值可以表示所有的字符了，所以哈希表大小定义为256.

字符为key，每个字符出现的次数为value

建立2个哈希表hash1、hash2，分别统计每个字符串出现的次数

最后比较hash1、hash2.

AC代码如下：

bool canConstruct(string ransomNote, string magazine)
{
if(ransomNote.length() > magazine.length())
return false;
int hash1[256] = {0};
int hash2[256] = {0};
for(int i = 0;i<ransomNote.size();i++)
hash1[ransomNote[i]]++;
for(int i = 0;i<magazine.size();i++)
hash2[magazine[i]]++;
for(int i =0;i<256;i++)
{
if(hash1[i] > hash2[i])
return false;
}
return true;
}