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
wordswill be in the range[1, 1000]. - The length of
words[i]will be in the range[1, 30].
排序+set解法
class Solution(object): def longestWord(self, words): """ :type words: List[str] :rtype: str """ # use greey algo # find the most length word that can be built one character at a time by other words in words words_set = set([""]) words.sort() ans = "" for word in words: if word[:-1] in words_set: if len(word) > len(ans): ans = word words_set.add(word) return ans
或者是trie：
class Node(object): def __init__(self, val=""): self.val = val self.subs = collections.defaultdict(Node)class Trie(object): def __init__(self): self.root = Node("") def insert(self, s): node = self.root for c in s: node = node.subs[c] node.val = s def longest_word(self): self.ans = "" def dfs(node): for k, n in node.subs.items(): if n.val: if len(n.val)>len(self.ans) or (len(n.val)==len(self.ans) and n.val
