Problem
Given an m x n board of characters and a list of strings words, return all words on the board.
Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
Algorithm
Use Trie for save and search word, run dfs find the word in the board.
Code
class Solution:def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:class Trie:def __init__(self):self.root = {}def insert(self, word):node = self.rootfor c in word:if c not in node:node[c] = {}node = node[c]node['leaf'] = word trie = Trie()for word in words:trie.insert(word)m, n = len(board), len(board[0])result = []def dfs(x, y, node):c = board[x][y]if c not in node:returnnext_node = node[c]word = next_node.get('leaf')if word:result.append(word)next_node['leaf'] = None # need removeboard[x][y] = '#'for dx, dy in [(-1,0), (1,0), (0,-1), (0,1)]:nx, ny = x + dx, y + dyif 0 <= nx < m and 0 <= ny < n and board[nx][ny] != '#':dfs(nx, ny, next_node)board[x][y] = cfor i in range(m):for j in range(n):dfs(i, j, trie.root)return list(result)