230313 자동완성

public static int solution(String[] words) {
    List<String> sortedWords = Arrays.asList(words);
    Collections.sort(sortedWords);

    int answer = len(sortedWords.get(0), sortedWords.get(1));
    answer = answer < sortedWords.get(0).length() ? answer + 1 : answer;

    for (int i = 1; i < sortedWords.size() - 1; i++) {
        int prevLen = len(sortedWords.get(i), sortedWords.get(i - 1));
        int nextLen = len(sortedWords.get(i), sortedWords.get(i + 1));
        int maxLength = Math.max(prevLen, nextLen);

        answer += maxLength < sortedWords.get(i).length() ? maxLength + 1 : maxLength;
    }

    int lastLen = len(sortedWords.get(sortedWords.size() - 2), sortedWords.get(sortedWords.size() - 1));
    answer += lastLen < sortedWords.get(sortedWords.size() - 1).length() ? lastLen + 1 : lastLen;

    return answer;
}

public static int len(String s1, String s2) {
    int length = 0;

    for (int i = 0; i < Math.min(s1.length(), s2.length()); i++) {
        if (s1.charAt(i) == s2.charAt(i)) {
            length++;
        } else {
            break;
        }
    }

    return length;
}

중복이 없으므로 정렬 후 인접한 값을 비교함으로써 답을 구할 수 있음