一、题目
题目很简单，输入一个字母组成的二维数组，以某一个字母为起点，向、上下左右搜索、练成的字符看是不是在给定的字符串数组中
二、解答
通过深度优先搜索，每一步和数组中的字符串匹配是可以计算出来正确结果的，但是结果超时
重点是匹配的过程时间太长
通过字典树，可以优化两点
一是检索某个字符串在不在的时间
一是一个某个字符串前缀的字符串是否存在于字符串数组中
最终的答案：
注意，结果去重
class Trie:
def __init__(self):
"""
Initialize your data structure here.
"""
self.x = {}
self.p = {}
def insert(self, word):
"""
Inserts a word into the trie.
:type word: str
:rtype: void
"""
if word:
self.x[word] = True
for i in range(1,len(word)+1):
self.p[word[0:i]] = True
def search(self, word):
"""
Returns if the word is in the trie.
:type word: str
:rtype: bool
"""
for x in self.x:
if word == x:
return True
return False
def startsWith(self, prefix):
"""
Returns if there is any word in the trie that starts with the given prefix.
:type prefix: str
:rtype: bool
"""
return prefix in self.p
class Solution:
def findWords(self, board, words):
"""
:type board: List[List[str]]
:type words: List[str]
:rtype: List[str]
"""
t = Trie()
for word in words:
t.insert(word)
visited = [([0 for j in range(len(board[0]))]) for i in range(len(board))]
res = []
for row in range(len(board)):
for col in range(len(board[0])):
self.DFS(board, visited, "", row, col, t, res)
return list(set(res))
def DFS(self, board, visited, temp_str, pos_row, pos_col, t, res):
if pos_row < 0 or pos_row >= len(board) or pos_col < 0 or pos_col >= len(board[0]):
return
# print(pos_row, pos_col, visited)
if visited[pos_row][pos_col] == 1:
return
temp_str += board[pos_row][pos_col]
if not t.startsWith(temp_str):
return
if t.search(temp_str):
res.append(temp_str)
visited[pos_row][pos_col] = 1
self.DFS(board, visited, temp_str, pos_row, pos_col + 1, t, res)
self.DFS(board, visited, temp_str, pos_row - 1, pos_col, t, res)
self.DFS(board, visited, temp_str, pos_row, pos_col - 1, t, res)
self.DFS(board, visited, temp_str, pos_row + 1, pos_col, t, res)
visited[pos_row][pos_col] = 0
if __name__ == "__main__":
s = Solution()
# print(s.findWords([
# ['o','a','a','n'],
# ['e','t','a','e'],
# ['i','h','k','r'],
# ['i','f','l','v']
# ],
# ["oath","pea","eat","rain"]))
print(s.findWords([["b"],["a"],["b"],["b"],["a"]],
["baa","abba","baab","aba"]))
三、参考
https://blog.csdn.net/ljiabin/article/details/45846527