leetcode解题: Ransom Note (383)

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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

解法1：O(N + M) Time with O(M) Space

经典的用hashtable解决的字母问题，把magazine先hash算出每一个字母出现的次数，然后对ransomNote扫描碰到见过的字母则次数-1，直到所对应的字母的次数小于0（false）或者是可以扫描完所有的ransomNote的字母。
C++

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class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> map (26);
        for (int i = 0; i < magazine.size(); ++i) {
            ++map[magazine[i] - 'a'];
        }
        for (int i = 0; i < ransomNote.size(); ++i) {
            if (map[ransomNote[i] - 'a'] <= 0) {
                return false;
            } else {
                --map[ransomNote[i] - 'a'];
            }
        }
        return true;
    }
};

C++ with std::unordered_map

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class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        unordered_map<char, int> map;
        for (int i = 0; i < magazine.size(); ++i) {
            map[magazine[i]]++;
        }
        for (int i = 0; i < ransomNote.size(); ++i) {
            if (map[ransomNote[i]] == 0) {
                return false;
            } else {
                --map[ransomNote[i]];
            }           
        }
        return true;
    }
};

Java

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