地址:https://leetcode-cn.com/problems/word-search-ii/
给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words，找出所有同时在二维网格和字典中出现的单词。
单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。
示例1:
输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]
示例2:
输入：board = [["a","b"],["c","d"]], words = ["abcb"]
输出：[]
没有解出来
class Solution:
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
def dfs(i, j, t, s):
ch = board[i][j]
if ch not in t:
return
t = t[ch]
if "end" in t and t["end"] == 1:
res.append(s+ch)
t["end"] = 0
board[i][j] = "#"
if i + 1 < m and board[i+1][j] != "#":
dfs(i+1, j, t, s+ch)
if i - 1 >= 0 and board[i-1][j] != "#":
dfs(i-1, j, t, s+ch)
if j + 1 < n and board[i][j+1] != "#":
dfs(i, j+1, t, s+ch)
if j - 1 >= 0 and board[i][j-1] != "#":
dfs(i, j-1, t, s+ch)
board[i][j] = ch
# 将word存入前缀树
trie = {}
for word in words:
t = trie
for ch in word:
if ch not in t:
t[ch] = {}
t = t[ch]
t["end"] = 1
m = len(board)
n = len(board[0])
res = []
# 对board进行深度优先搜索
for i in range(m):
for j in range(n):
dfs(i, j, trie, "")
return res