LeetCode 212. Word Search II 5行内Build Trie树

本文链接：https://blog.youkuaiyun.com/taoqick/article/details/105633229

本文介绍了一种在二维字符板上寻找字典中所有单词的算法。通过深度优先搜索(DFS)结合字典树(Trie)，实现了高效查找。文章详细展示了如何构建字典树并进行搜索的过程。

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Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example:

Input: 
board = [
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]
words = ["oath","pea","eat","rain"]

Output: ["eat","oath"]

Note:

All inputs are consist of lowercase letters a-z.
The values of words are distinct.

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If defining a Trie class, it needs mutiple lines. But if using a nested dict, the codes look quite clean:

class Solution:
    def dfs(self, board, x, y, rt, rows, cols, res):
        cur_node, cur_ch = rt[board[x][y]], board[x][y]
        if ("$" in cur_node):
            res.append(cur_node["$"])
            cur_node.pop("$")
        board[x][y] = '#'
        for (dx, dy) in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
            nx, ny = x + dx, y + dy
            if (nx < rows and nx >= 0 and ny < cols and ny >= 0 and board[nx][ny] != '#' and board[nx][ny] in cur_node):
                self.dfs(board, nx, ny, cur_node, rows, cols, res)
        board[x][y] = cur_ch

    def findWords(self, board, words):
        trie, res = {}, []
        for word in words:
            node = trie
            for ch in word:
                node = node.setdefault(ch, {})
            node["$"] = word
        print(trie)
        rows, cols = len(board), len(board[0]) if len(board) > 0 else 0
        if (cols == 0):
            return []
        for i in range(rows):
            for j in range(cols):
                if (board[i][j] in trie):
                    self.dfs(board, i, j, trie, rows, cols, res)
        return res

s = Solution()
board = [
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]
words = ["oath","pea","eat","rain"]
print(s.findWords(board, words))