题目：原题链接（困难）
标签：回溯算法、字典树
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( W × L 2 ) O(W×L^2) O(W×L2) | O ( W × L 2 ) O(W×L^2) O(W×L2) | 264ms (97.67%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
def build_tree(words):
tree = {}
for i, word in enumerate(words):
node = tree
for ch in word:
if ch not in node:
node[ch] = {}
node = node[ch]
node["@"] = i
return tree
class Solution:
def __init__(self):
self.words = []
self.size = 0
self.tree = {}
self.dic = collections.defaultdict(list)
def wordSquares(self, words: List[str]) -> List[List[str]]:
self.words = words
self.size = len(words[0])
# 构造字典树
self.tree = build_tree(words)
# 构造开头字母对应字典
for word in words:
for i in range(self.size):
self.dic[word[0:i + 1]].append(word)
ans = []
for word in words:
ans.extend(self.check(word))
ans.sort()
return ans
def check(self, word):
"""检查指定词语"""
nodes = []
for i in range(self.size):
node = self.tree
nodes.append(node)
find = True
for i in range(self.size):
if word[i] in nodes[i]:
nodes[i] = nodes[i][word[i]]
else:
find = False
break
if not find:
return []
return self.count([word], word, nodes)
def count(self, path, word1, nodes):
"""寻找下一行的词语"""
idx = len(path)
# 处理递归完成的情况
if idx == self.size:
return [path]
# 寻找已经被确定的字符
confirm = []
for i in range(idx):
confirm.append(path[i][idx])
confirm = "".join(confirm)
ans = []
for word2 in self.dic[confirm]:
new_nodes = []
for i in range(self.size):
if word2[i] in nodes[i]:
new_nodes.append(nodes[i][word2[i]])
else:
break
else:
ans.extend(self.count(path + [word2], word1, new_nodes))
return ans
