132. Word Search II
Chinese
English
Given a matrix and a dictionary of lowercase letters. Find all the words also appear in the dictionary and matrix. A word can begin anywhere in the matrix, you can be left / right / upper / lower four adjacent moving direction. A letter in a word can only be used once. And there is no repetition of words in the dictionary
Sample
Example 1:
输入：["doaf","agai","dcan"]，["dog","dad","dgdg","can","again"]
输出：["again","can","dad","dog"]
解释：
d o a f
a g a i
d c a n
矩阵中查找，返回 ["again","can","dad","dog"]。
Sample 2:
输入：["a"]，["b"]
输出：[]
解释：
a
矩阵中查找，返回 []。
challenge
Use trie to implement your algorithm
DIRECTIONS = [(0, -1), (0, 1), (-1, 0), (1, 0)]
class Solution:
"""
@param board: A list of lists of character
@param words: A list of string
@return: A list of string
"""
def wordSearchII(self, board, words):
if board is None or len(board) == 0:
return []
word_set = set(words)
prefix_set = set()
for word in words:
for i in range(len(word)):
prefix_set.add(word[:i + 1])
result = set()
for i in range(len(board)):
for j in range(len(board[0])):
c = board[i][j]
self.search(
board,
i,
j,
board[i][j],
word_set,
prefix_set,
set([(i, j)]),
result,
)
return list(result)
def search(self, board, x, y, word, word_set, prefix_set, visited, result):
if word not in prefix_set:
return
if word in word_set:
result.add(word)
for delta_x, delta_y in DIRECTIONS:
x_ = x + delta_x
y_ = y + delta_y
if not self.inside(board, x_, y_):
continue
if (x_, y_) in visited:
continue
visited.add((x_, y_))
self.search(
board,
x_,
y_,
word + board[x_][y_],
word_set,
prefix_set,
visited,
result,
)
visited.remove((x_, y_))
def inside(self, board, x, y):
return 0 <= x < len(board) and 0 <= y < len(board[0])
Note the use of prefix hash map to pruning.
Solution trie use of:
= ThereDIRECTIONS [(0, -1), (0,. 1), (-1, 0), (. 1, 0)]
class TrieNode: #define trie nodes
DEF the __init __ (Self):
self.children = {}
= False self.is_word
self.word = None
class Trie:
DEF __init __ (Self):
self.root = TrieNode ()
DEF the Add (Self, Word): # trie insert the word
the Node = self.root
for c in Word:
IF node.children not in c:
node.children [c] = TrieNode () # application at this node node
# = node.children continue to traverse the node [c]
node.is_word = True
node.word = # into Word word
def find (Self, Word):
Node = self.root
for c in Word:
Node = node.children.get (C)
IF None Node IS:
return None
return Node
class Solution:
"" "
@param Board: A Character List of Lists of
@param words: A List of String
@return : List of String A
"" "
DEF wordSearchII (Self, Board, words):
IF Board or IS None len (Board) == 0:
return []
Trie Trie = ()
for Word in words: # insert word
trie.add (Word)
the Result = the SET ()
for i in the Range (len (Board)): # ergodic matrix alphabet, each letter as the first letter of the word to start the search
for j in range(len(board[0])):
c = board[i][j]
self.search(
board,
i,
j,
trie.root.children.get(c),
set([(i, j)]),
result,
)
return list(result)
def search(self, board, x, y, node, visited, result): #在字典树上dfs查找
if node is None:
return
if node.is_word:
result.add(node.word)
for delta_x, delta_y in DIRECTIONS: #向四个方向查找
x_ = x + delta_x
y_ = y + delta_y
if not self.inside(board, x_, y_):
continue
if (x_, y_) in visited:
continue
visited.add((x_, y_))
self.search(
board,
x_,
y_,
node.children.get(board[x_][y_]),
visited,
result,
)
visited.remove((x_, y_))
def inside(self, board, x, y):
return 0 <= x < len(board) and 0 <= y < len(board[0])