LintCode: Longest Words


LintCode: Longest Words
Given a dictionary, find all of the longest words in the dictionary.
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?
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。
https://codesolutiony.wordpress.com/2015/06/07/lintcode-longest-words/
    ArrayList<String> longestWords(String[] dictionary) {
        ArrayList<String> res = new ArrayList<String>();
        for (String str : dictionary) {
            if (res.isEmpty() || res.get(0).length() < str.length()) {
                res.clear();
                res.add(str);
            } else if (res.get(0).length() == str.length()) {
                res.add(str);
            }
        }
        return res;
    }

    ArrayList<String> longestWords(String[] dictionary) {
        // write your code here
        ArrayList<String> result = new ArrayList<String>();
        int maxLen = 0;
        for(String word : dictionary){
            if(word.length() > maxLen){
                result.clear();
                result.add(word);
                maxLen = word.length();
            } else if(word.length() == maxLen){
                result.add(word);
            }
        }
        return result;
    }
http://www.chenguanghe.com/lintcode-longest-words/
ArrayList<String> longestWords(String[] dictionary) {
        // write your code here
        ArrayList<String> res = new ArrayList<String>();
        if(dictionary == null || dictionary.length == 0)
            return res;
        int len = dictionary[0].length();
        for(int i = 0; i < dictionary.length; i++) {
            if(dictionary[i].length() == len)
                res.add(dictionary[i]);
            else if(dictionary[i].length() > len){
                len = dictionary[i].length();
                res.clear();
                res.add(dictionary[i]);
            }
        }
        return res;
    }
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