QUESTION:

Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.

If there is no answer, return the empty string. Example 1: Input: words = [“w”,”wo”,”wor”,”worl”, “world”] Output: “world” Explanation: The word “world” can be built one character at a time by “w”, “wo”, “wor”, and “worl”. Example 2: Input: words = [“a”, “banana”, “app”, “appl”, “ap”, “apply”, “apple”] Output: “apple” Explanation: Both “apply” and “apple” can be built from other words in the dictionary. However, “apple” is lexicographically smaller than “apply”. Note:

All the strings in the input will only contain lowercase letters. The length of words will be in the range [1, 1000]. The length of words[i] will be in the range [1, 30].

EXPLANATION：

首先这个题目的意思是给一串字符串，能从第一个字母拼到最后一个字母，最长能拼到字符是。 算法是： 1.将所有的字符都放在set中 2.进行排序，最长的放在后面 3.进行反向查找，先拿出最长的字符，看是否满足每个substring都存在。

SOLUTION：

class Solution {
    public String longestWord(String[] words) {
        Set<String> wordset = new HashSet();
        for (String word: words) wordset.add(word);
        Arrays.sort(words, (a, b) -> a.length() == b.length()
                ? a.compareTo(b) : b.length() - a.length());
        for (String word: words) {
            boolean good = true;
            for (int k = 1; k < word.length(); ++k) {
                if (!wordset.contains(word.substring(0, k))) {
                    good = false;
                    break;
                }
            }
            if (good) return word;
        }
        return "";
    }
}