［LeetCode］383. Ransom Note 解题小结

Posted 2020-08-09 医生工程师

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;
    }
};