C#LeetCode刷题之#720-词典中最长的单词(Longest Word in Dictionary)

问题

该文章的最新版本已迁移至个人博客【比特飞】,单击链接 https://www.byteflying.com/archives/4120 访问。

给出一个字符串数组words组成的一本英语词典。从中找出最长的一个单词,该单词是由words词典中其他单词逐步添加一个字母组成。若其中有多个可行的答案,则返回答案中字典序最小的单词。

若无答案,则返回空字符串。

输入: words = ["w","wo","wor","worl", "world"]

输出: "world"

解释: 单词"world"可由"w", "wo", "wor", 和 "worl"添加一个字母组成。

输入: words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]

输出: "apple"

解释: "apply"和"apple"都能由词典中的单词组成。但是"apple"得字典序小于"apply"。

注意:

所有输入的字符串都只包含小写字母。
words数组长度范围为[1,1000]。
words[i]的长度范围为[1,30]。


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.

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".

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].


示例

该文章的最新版本已迁移至个人博客【比特飞】,单击链接 https://www.byteflying.com/archives/4120 访问。

public class Program {

    public static void Main(string[] args) {
        var words = new string[] { "a", "banana", "app", "appl", "ap", "apply", "apple" };

        var res = LongestWord(words);
        Console.WriteLine(res);

        Console.ReadKey();
    }

    private static string LongestWord(string[] words) {
        var wordsSet = new HashSet();
        var resultSet = new HashSet();
        Array.Sort(words);
        for(var i = 0; i < words.Length; i++) {
            wordsSet.Add(words[i]);
        }
        for(var i = words.Length - 1; i >= 0; i--) {
            if(IsCompleteWord(words[i], wordsSet)) {
                resultSet.Add(words[i]);
            }
        }
        var list = resultSet.OrderByDescending(r => r.Length).ToList();
        return list.Where(r => r.Length == list[0].Length).Min();
    }

    private static bool IsCompleteWord(string word, HashSet wordsSet) {
        for(var i = 0; i < word.Length; i++) {
            if(!wordsSet.Contains(word.Substring(0, i + 1))) return false;
        }
        return true;
    }

}

以上给出1种算法实现,以下是这个案例的输出结果:

该文章的最新版本已迁移至个人博客【比特飞】,单击链接 https://www.byteflying.com/archives/4120 访问。

apple

分析:

设单词的最大长度为m,那么以上算法的时间复杂度应为: O(m*n) 。

你可能感兴趣的:(C#LeetCode刷题,C#LeetCode)