LintCode: Longest Words
Challenge
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"].
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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