给定一个单词集合 words （没有重复），找出并返回其中所有的 单词方块 。 words 中的同一个单词可以被 多次 使用。你可以按 任意顺序 返回答案。
一个单词序列形成了一个有效的 单词方块 的意思是指从第 k 行和第 k 列 0 <= k < max(numRows, numColumns) 来看都是相同的字符串。
- 例如，单词序列
["ball","area","lead","lady"]形成了一个单词方块，因为每个单词从水平方向看和从竖直方向看都是相同的。
示例 1：
输入：words = ["area","lead","wall","lady","ball"] 输出: [["ball","area","lead","lady"], ["wall","area","lead","lady"]] 解释： 输出包含两个单词方块，输出的顺序不重要，只需要保证每个单词方块内的单词顺序正确即可。
示例 2：
输入：words = ["abat","baba","atan","atal"] 输出：[["baba","abat","baba","atal"], ["baba","abat","baba","atan"]] 解释： 输出包含两个单词方块，输出的顺序不重要，只需要保证每个单词方块内的单词顺序正确即可。
提示:
1 <= words.length <= 10001 <= words[i].length <= 4words[i]长度相同words[i]只由小写英文字母组成words[i]都 各不相同
根据已添加单词确定下一个单词的前缀，继续进行搜索。
比如已经添加了两个单词
class Trie:
def __init__(self):
self.children = [None] * 26
self.v = []
def insert(self, w, i):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.v.append(i)
def search(self, w):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return []
node = node.children[idx]
return node.v
class Solution:
def wordSquares(self, words: List[str]) -> List[List[str]]:
def dfs(t):
if len(t) == len(words[0]):
ans.append(t[:])
return
idx = len(t)
pref = [v[idx] for v in t]
indexes = trie.search(''.join(pref))
for i in indexes:
t.append(words[i])
dfs(t)
t.pop()
trie = Trie()
ans = []
for i, w in enumerate(words):
trie.insert(w, i)
for w in words:
dfs([w])
return ansclass Trie {
Trie[] children = new Trie[26];
List<Integer> v = new ArrayList<>();
void insert(String word, int i) {
Trie node = this;
for (char c : word.toCharArray()) {
c -= 'a';
if (node.children[c] == null) {
node.children[c] = new Trie();
}
node = node.children[c];
node.v.add(i);
}
}
List<Integer> search(String pref) {
Trie node = this;
for (char c : pref.toCharArray()) {
c -= 'a';
if (node.children[c] == null) {
return Collections.emptyList();
}
node = node.children[c];
}
return node.v;
}
}
class Solution {
private Trie trie = new Trie();
private String[] words;
private List<List<String>> ans = new ArrayList<>();
public List<List<String>> wordSquares(String[] words) {
this.words = words;
for (int i = 0; i < words.length; ++i) {
trie.insert(words[i], i);
}
List<String> t = new ArrayList<>();
for (String w : words) {
t.add(w);
dfs(t);
t.remove(t.size() - 1);
}
return ans;
}
private void dfs(List<String> t) {
if (t.size() == words[0].length()) {
ans.add(new ArrayList<>(t));
return;
}
int idx = t.size();
StringBuilder pref = new StringBuilder();
for (String x : t) {
pref.append(x.charAt(idx));
}
List<Integer> indexes = trie.search(pref.toString());
for (int i : indexes) {
t.add(words[i]);
dfs(t);
t.remove(t.size() - 1);
}
}
}type Trie struct {
children [26]*Trie
v []int
}
func newTrie() *Trie {
return &Trie{}
}
func (this *Trie) insert(word string, i int) {
node := this
for _, c := range word {
c -= 'a'
if node.children[c] == nil {
node.children[c] = newTrie()
}
node = node.children[c]
node.v = append(node.v, i)
}
}
func (this *Trie) search(word string) []int {
node := this
for _, c := range word {
c -= 'a'
if node.children[c] == nil {
return []int{}
}
node = node.children[c]
}
return node.v
}
func wordSquares(words []string) [][]string {
trie := newTrie()
for i, w := range words {
trie.insert(w, i)
}
ans := [][]string{}
var dfs func([]string)
dfs = func(t []string) {
if len(t) == len(words[0]) {
cp := make([]string, len(t))
copy(cp, t)
ans = append(ans, cp)
return
}
idx := len(t)
pref := []byte{}
for _, v := range t {
pref = append(pref, v[idx])
}
indexes := trie.search(string(pref))
for _, i := range indexes {
t = append(t, words[i])
dfs(t)
t = t[:len(t)-1]
}
}
for _, w := range words {
dfs([]string{w})
}
return ans
}