LeetCode 720. Longest Word in Dictionary

原题链接在这里：https://leetcode.com/problems/longest-word-in-dictionary/description/

题目：

Given an array of strings words representing an English Dictionary, return 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.

Note that the word should be built from left to right with each additional character being added to the end of a previous word. 

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".

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 30
• words[i] consists of lowercase English letters.

题解：

Sort the words, then with the same prefix, longer words will come after shorter words.

Iterate the words, for current word, if its prefix already exist in the visited set, we can add it to the visited set and update the result.

Time Complexity: O(nlogn + n * len). n = words.length. len is the length of longest length of word.

Space: O(n * len).

AC Java:

class Solution {
    public String longestWord(String[] words) {
        String res = "";
        if(words == null || words.length == 0){
            return res;
        }
        
        Arrays.sort(words);
        HashSet<String> visited = new HashSet<>();
        for(String w : words){
            if(w.length() == 1 || visited.contains(w.substring(0, w.length() - 1))){
                visited.add(w);
                if(w.length() > res.length()){
                    res = w;
                }
            }
        }
        
        return res;
    }
}