383. Ransom Note

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

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;
}
};