题目大意
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给定一组单词 words。
要求：找出其中的最长单词，且该单词由这组单词中的其他单词组合而成。若有多个长度相同的结果，返回其中字典序最小的一项，若没有符合要求的单词则返回空字符串。
解题思路
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先将所有单词按照长度从长到短排序，相同长度的字典序小的排在前面。然后将所有单词存入字典树中。
然后一重循环遍历所有单词 word，二重循环遍历单词中所有字符 word[i]。
如果当前遍历的字符为单词末尾，递归判断从 i + 1 位置开始，剩余部分是否可以切分为其他单词组合，如果可以切分，则返回当前单词 word。如果不可以切分，则返回空字符串 ""。
代码
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class Trie:
def __init__(self):
"""
Initialize your data structure here.
"""
self.children = dict()
self.isEnd = False
def insert(self, word: str) -> None:
"""
Inserts a word into the trie.
"""
cur = self
for ch in word:
if ch not in cur.children:
cur.children[ch] = Trie()
cur = cur.children[ch]
cur.isEnd = True
def search(self, word: str) -> bool:
"""
Returns if the word is in the trie.
"""
cur = self
for ch in word:
if ch not in cur.children:
return False
cur = cur.children[ch]
return cur is not None and cur.isEnd
def splitToWord(self, remain):
if not remain or remain == "":
return True
cur = self
for i in range(len(remain)):
ch = remain[i]
if ch not in cur.children:
return False
if cur.children[ch].isEnd and self.splitToWord(remain[i + 1:]):
return True
cur = cur.children[ch]
return False
def dfs(self, words):
for word in words:
cur = self
size = len(word)
for i in range(size):
ch = word[i]
if i < size - 1 and cur.children[ch].isEnd and self.splitToWord(word[i+1:]):
return word
cur = cur.children[ch]
return ""
class Solution:
def longestWord(self, words: List[str]) -> str:
words.sort(key=lambda x: (-len(x), x))
trie_tree = Trie()
for word in words:
trie_tree.insert(word)
ans = trie_tree.dfs(words)
return ans
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