题目:
?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
建立字符表，对应字符的个数，如果ransom Note的字符数大于magazine的，返回false。
class Solution { public: bool canConstruct(string ransomNote, string magazine) { if(ransomNote.length() > magazine.length()) return false; int dictRansom[256]={0}; int dictMagazine[256]={0}; for(int i = 0; i < ransomNote.length(); ++i){ dictRansom[ransomNote[i]]++; } for(int i = 0; i < magazine.length(); ++i){ dictMagazine[magazine[i]]++; } for(int i = 0; i < ransomNote.length(); ++i){ if(dictRansom[ransomNote[i]] > dictMagazine[ransomNote[i]]) return false; } return true; } };
［LeetCode］383. Ransom Note 解题小结
原文：http://www.cnblogs.com/Doctengineer/p/5883748.html