[leetcode] 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

这道题是判断给定字符串能否由另一字符串中字符组合生成，题目难度为Easy。

通过Hash Table统计magazine字符串中各字符出现的次数，然后遍历ransomNote字符串，每遍历一个字符在Hash Table中将其计数减一，如果计数为负则返回false，如果顺利遍历完整个ransomNote字符串则返回true。由于字符只有26种，因此可以用vector来取代unordered_map实现Hash
Table的功能，一定程度上提高执行效率。具体代码：class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
vector<int> hash(26, 0);
for(auto c:magazine) ++hash[c-'a'];
for(auto c:ransomNote) {
if(--hash[c-'a'] < 0) return false;
}
return true;
}
};