# Longest Words ### Source - lintcode: [(133) Longest Words](http://www.lintcode.com/en/problem/longest-words/) ~~~ Given a dictionary, find all of the longest words in the dictionary. Example Given { "dog", "google", "facebook", "internationalization", "blabla" } the longest words are(is) ["internationalization"]. Given { "like", "love", "hate", "yes" } the longest words are ["like", "love", "hate"]. Challenge It's easy to solve it in two passes, can you do it in one pass? ~~~ ### 题解 简单题，容易想到的是首先遍历以便，找到最长的字符串，第二次遍历时取最长的放到最终结果中。但是如果只能进行一次遍历呢？一次遍历意味着需要维护当前遍历的最长字符串，这必然有比较与更新删除操作，这种情况下使用双端队列最为合适，这道题稍微特殊一点，不必从尾端插入，只需在遍历时若发现比数组中最长的元素还长时删除整个列表。 ### Java ~~~ class Solution { /** * @param dictionary: an array of strings * @return: an arraylist of strings */ ArrayList<String> longestWords(String[] dictionary) { ArrayList<String> result = new ArrayList<String>(); if (dictionary == null || dictionary.length == 0) return result; for (String str : dictionary) { // combine empty and shorter length if (result.isEmpty() || str.length() > result.get(0).length()) { result.clear(); result.add(str); } else if (str.length() == result.get(0).length()) { result.add(str); } } return result; } } ~~~ ### 源码分析 熟悉变长数组的常用操作。 ### 复杂度分析 时间复杂度 O(n)O(n)O(n), 最坏情况下需要保存 n - 1个字符串，空间复杂度 O(n)O(n)O(n). ### Reference - [Lintcode: Longest Words | codesolutiony](https://codesolutiony.wordpress.com/2015/06/07/lintcode-longest-words/)