原题
给出一个由小写字母组成的矩阵和一个字典。找出所有同时在字典和矩阵中出现的单词。一个单词可以从矩阵中的任意位置开始，可以向左/右/上/下四个相邻方向移动。
样例
给出矩阵：
doaf
agai
dcan
和字典：
{"dog", "dad", "dgdg", "can", "again"}
返回 {"dog", "dad", "can", "again"}
解题思路
- 最直接但是会超时的方法 - 先把字典做成hashMap，对每一个点DFS，每增加一个字符就去hashMap查找是否存在。一共m*n个点，每次DFS复杂度是O(m*n)，所以总的时间复杂度是O(m*n * m*n)
- 可行解 - 把给出的字典变成Trie树，Trie树可以检查前缀，hashMap做不到。检查前缀的时候，如果发现某个前缀不存在，就可以不用继续DFS了，相当于剪枝
- 注意：当不必要的check增多时，会导致TLE
# 超时
if x < 0 or x >= len(board) or y < 0 or y >= len(board[0]) or board[x][y] == 0 or cur_node == None:
# 免去两个不必要的check，不超时
if x < 0 or x >= len(board) or y < 0 or y >= len(board[0]):
完整代码
class TrieNode(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.children = {}
self.IsWord = False
class Trie(object):
def __init__(self):
self.root = TrieNode()
def insert(self, word):
"""
Inserts a word into the trie.
:type word: str
:rtype: void
"""
node = self.root
for letter in word:
child = node.children.get(letter)
if child is None:
child = TrieNode()
node.children[letter] = child
node = child
node.IsWord = True
class Solution(object):
def __init__(self):
self.result = []
def findWords(self, board, words):
"""
:type board: List[List[str]]
:type words: List[str]
:rtype: List[str]
"""
trie = Trie()
for word in words:
trie.insert(word)
for row_num in range(len(board)):
for col_num in range(len(board[0])):
self.search(board, row_num, col_num, trie.root, "")
return self.result
def search(self, board, x, y, cur_node, word):
if cur_node.IsWord:
self.result.append(word)
cur_node.IsWord = False
if not (x < 0 or x >= len(board) or y < 0 or y >= len(board[0])):
temp = board[x][y]
cur_node = cur_node.children.get(temp)
if cur_node:
board[x][y] = "#"
self.search(board, x+1, y, cur_node, word+temp)
self.search(board, x-1, y, cur_node, word+temp)
self.search(board, x, y+1, cur_node, word+temp)
self.search(board, x, y-1, cur_node, word+temp)
board[x][y] = temp
