[LeetCode] 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].
class Solution {
public:
    string longestWord(vector<string>& words) {
        sort(words.begin(), words.end());
        unordered_set<string> ss;
        string res;
        for (string word : words) {
            if (word.size() == 1 || ss.count(word.substr(0, word.size() - 1))) {
                res = word.size() > res.size() ? word : res;
                ss.insert(word);
            }
        }
        return res;
    }
};
// 46 ms
时间: 2024-11-10 05:31:06

[LeetCode] Longest Word In Dictionary的相关文章

[LeetCode] Longest Word in Dictionary through Deleting 删除后得到的字典中的最长单词

Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographic

leetcode 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 l

[LeetCode] 524. Longest Word in Dictionary through Deleting

Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographic

Hash Table-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 l

[leetcode]720. Longest Word in Dictionary字典中最长的单词

b.compareTo(a) 这个函数是比较两个值得大小,如果b比a大,那么返回1 如果小,那么返回-1,相等返回0 如果比较的是字符串,那么比较字典编纂顺序,b靠前返回-1,靠后返回1 这个题的核心虽然是hashtable,但是这个方法还是很重要的,因为自己实现比较字符串很麻烦 /* 用一个Hashset来判断子串是不是在集合里 然后遍历出最长的就行 长度相同的用compareTo方法判断谁的字典顺序靠前 这个题最难的感觉其实是不知道compareTo方法,自己实现还挺麻烦的 */ publi

524. Longest Word in Dictionary through Deleting

https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/#/solutions An alternate, more efficient solution which avoids sorting the dictionary: public String findLongestWord(String s, List<String> d) { String longest = "";

[LeetCode] 676. Implement Magic Dictionary 实现神奇字典

Implement a magic directory with buildDict, and search methods. For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary. For the method search, you'll be given a word, and judge whether if you modify exactly one

【leetcode】Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, givens = "catsanddog",dict = ["cat", "cats"

【leetcode】Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &