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].
本质上是trie！前缀树！
class Solution(object):
def longestWord(self, words):
"""
:type words: List[str]
:rtype: str
"""
# use greey algo
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
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<self.ans): self.ans = n.val dfs(n) dfs(self.root) return self.ans class Solution(object): def longestWord(self, words): """ :type words: List[str] :rtype: str """ trie = Trie() for word in words: trie.insert(word) return trie.longest_word()
注意：Trie每个结点的子树的根节点的组织方式有几种。
1>如果默认包含所有字符集，则查找速度快但浪费空间（特别是靠近树底部叶子）。
2>如果用链接法(如左儿子右兄弟)，则节省空间但查找需顺序（部分）遍历链表。
3>alphabet reduction: 减少字符宽度以减少字母集个数。
4>对字符集使用bitmap，再配合链接法。
Trie的插入和查找算法：
def find(node, key):
for char in key:
if char in node.children:
node = node.children[char]
else:
return None
return node.value == key
algorithm insert(root : node, s : string, value : any):
node = root
i = 0
n = length(s)
while i < n:
if node.child(s[i]) != nil:
node = node.child(s[i])
i = i + 1
else:
break
(* append new nodes, if necessary *)
while i < n:
node.child(s[i]) = new node
node = node.child(s[i])
i = i + 1
node.value = value
class TrieNode(object): def __init__(self): self.children=collections.defaultdict(TrieNode) self.isEnd=False self.word ='' class Trie(object): def __init__(self): self.root=TrieNode() def insert(self, word): node=self.root for c in word: node =node.children[c] node.isEnd=True node.word=word def bfs(self): q=collections.deque([self.root]) res='' while q: cur=q.popleft() for n in cur.children.values(): if n.isEnd: q.append(n) if len(n.word)>len(res) or n.word<res: res=n.word return res class Solution(object): def longestWord(self, words): trie = Trie() for w in words: trie.insert(w) return trie.bfs()
java的解法：
Build a trie in the normal way, then do a dfs to find the longest continuous downward path from the root. This is not a particularly hard question in the context of trie, the point of this solution is to present a generic way of trie building and inserting that can be easily adapted to similar questions. Code:
class Solution {
public String longestWord(String[] words) {
TrieNode root = new TrieNode ();
root.word = "-";
for (String word : words)
root.insert (word);
return dfs (root, "");
}
String dfs (TrieNode node, String accum) {
if (node == null || node.word.length () == 0)
return accum;
String res = "";
if (!node.word.equals ("-"))
accum = node.word;
for (TrieNode child : node.links) {
String curRes = dfs (child, accum);
if (curRes.length () > res.length () || (curRes.length () == res.length () && curRes.compareTo (res) < 0))
res = curRes;
}
return res;
}
/* Hand write this class every time you need to so you can remember well */
static class TrieNode {
String word = "";
TrieNode[] links = new TrieNode[26];
void insert (String s) {
char[] chs = s.toCharArray ();
TrieNode curNode = this;
for (int i = 0; i < chs.length; i++) {
int index = chs[i] - 'a';
if (curNode.links[index] == null)
curNode.links[index] = new TrieNode ();
curNode = curNode.links[index];
}
curNode.word = s;
}
}
}
A typical trie for the list of "ab", "ac":
class Solution {
class TrieNode {
TrieNode[] children;
boolean isWord;
String word;
public TrieNode() {
children = new TrieNode[26];
}
}
class Trie {
private TrieNode root;
public Trie() {
root = new TrieNode();
}
public void insert(String word) {
TrieNode node = root;
for (int i = 0; i < word.length(); i++) {
int idx = word.charAt(i) - 'a';
if (node.children[idx] == null) {
node.children[idx] = new TrieNode();
}
node = node.children[idx];
}
node.isWord = true;
node.word = word;
}
public String findLongestWord() {
String result = null;
Queue<TrieNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TrieNode node = queue.poll();
for (int j = 25; j >= 0; j--) {
if (node.children[j] != null && node.children[j].isWord) {
result = node.children[j].word;
queue.offer(node.children[j]);
}
}
}
}
return result;
}
}
public String longestWord(String[] words) {
Trie trie = new Trie();
for (String word : words) {
trie.insert(word);
}
return trie.findLongestWord();
}
}