letecode [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

题目大意：

　　判断字符串1能否由字符串2中的字符构成。

理  解：

　　构建数组记录字符串str2字符的数量，遍历str1，每遍历一个字符，数量减一，当数量减至-1时，返回false.

代 码 C++：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int hash[26];
        for(int i=0;i<26;++i){
            hash[i]=0;
        }
        for(char v1:magazine){
            hash[v1-'a']++;
        }
        for(char v2:ransomNote){
            hash[v2-'a']--;
            if(hash[v2-'a']==-1)
                return false;
        }
        return true;
    }
};

运行结果：

　　执行用时 :20 ms, 在所有 C++ 提交中击败了95.99%的用户

　　内存消耗 :11 MB, 在所有 C++ 提交中击败了80.00%的用户