720. Longest Word in Dictionary

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].

找出数组里最长的字符串，该字符串的子集也要在这个数组里，有多个答案则输出字典序最小的

C++(52ms):

class Solution {
public:
    string longestWord(vector<string>& words) {
        sort(words.begin(),words.end()) ;
        unordered_set<string> Set;
        string res = "" ;
        for(string w : words){
            if (w.length() == 1 || Set.count(w.substr(0,w.length()-1))){
                res = w.length() > res.length() ? w : res ;
                Set.insert(w) ;
            }
        }
        return res ;     
    }
};

Java(36ms):

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