Leetcode 212. Word Search II

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.root
                for 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:
                return
            next_node = node[c]
            word = next_node.get('leaf')
            if word:
                result.append(word)
                next_node['leaf'] = None # need remove

            board[x][y] = '#'
            for dx, dy in [(-1,0), (1,0), (0,-1), (0,1)]:
                nx, ny = x + dx, y + dy
                if 0 <= nx < m and 0 <= ny < n and board[nx][ny] != '#':
                    dfs(nx, ny, next_node)
            board[x][y] = c

        for i in range(m):
            for j in range(n):
                dfs(i, j, trie.root)

        return list(result)

原文地址：https://blog.csdn.net/mobius_strip/article/details/146445156

免责声明：本站文章内容转载自网络资源，如侵犯了原著者的合法权益，可联系本站删除。更多内容请关注自学内容网（zxcms.com）！