Leetcode: hash table 06 (ransom letter)

383. Ransom letter

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Give a ransom (ransom) String and a magazine (magazine) character string , Judge the first string ransom Can I have a second string magazines The characters inside make up . If it can be constructed , return true ; Otherwise return to false.

( Title Description ： In order not to reveal the handwriting of the ransom letter , To search magazines for the letters you need , Make up words to express meaning . Each character in the magazine string can only be used once in the ransom letter string .)

    /** *  Here are two strings ：ransomNote  and  magazine , Judge  ransomNote  Can it be done by  magazine  The characters inside make up . *  If possible , return  true ; Otherwise return to  false . * magazine  Each character in can only be in  ransomNote  Used once in . * <p> *  This question is to ask whether two words are heteronymous letters . * * @param ransomNote * @param magazine * @return */
    public static boolean canConstruct(String ransomNote, String magazine) {
    
        final byte[] magazineBytes = magazine.getBytes();
        HashMap<Byte, Integer> map = new HashMap<>();
        for (byte i : magazineBytes) {
    
            // If not , add to , Include quantity +1
            if (map.containsKey(i)) {
    
                map.put(i, map.get(i) + 1);
            } else {
    
                map.put(i, 1);
            }
        }
        final byte[] ransomNoteBytes = ransomNote.getBytes();
        int value;
        for (byte i : ransomNoteBytes) {
    
            // If not , be false
            if (!map.containsKey(i)) {
    
                return false;
            } else {
    
                // If there is only one last chance to use this letter , Then delete this key 
                value = map.get(i);
                if (value == 1) {
    
                    map.remove(i);
                } else {
    
                    map.put(i, value - 1);
                }
                if (map.size() == 0) {
    
                    return true;
                }
            }
        }
        return true;
    }

It's very similar to the same question , But not all , As long as the right contains the left completely , Need to count .
Ultimate idea ： Use arrays to do , It looks a lot simpler , Although the idea is the same .

    /** *  Here are two strings ：ransomNote  and  magazine , Judge  ransomNote  Can it be done by  magazine  The characters inside make up . *  If possible , return  true ; Otherwise return to  false . * magazine  Each character in can only be in  ransomNote  Used once in . * <p> *  This question is to ask whether two words are heteronymous letters . * * @param ransomNote * @param magazine * @return */
    public static boolean canConstruct(String ransomNote, String magazine) {
    
        final byte[] magazineBytes = magazine.getBytes();
        int[] result = new int[26];
        for (byte i : magazineBytes) {
    
            result[i - 'a'] += 1;
        }
        final byte[] ransomNoteBytes = ransomNote.getBytes();
        for (byte i : ransomNoteBytes) {
    
            if (result[i - 'a'] <= 0) {
    
                return false;
            }else {
    
                result[i - 'a'] -= 1;
            }
        }
        return true;
    }