LeetCode 212. Word Search II

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题目

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:

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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.

思路

Trie + DFS.

代码

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class TrieNode():
    def __init__(self):
        self.children = collections.defaultdict(TrieNode)
        self.isWord = False
    
class Trie():
    def __init__(self):
        self.root = TrieNode()
    
    def insert(self, word):
        node = self.root
        for w in word:
            node = node.children[w]
        node.isWord = True
    
    def search(self, word):
        node = self.root
        for w in word:
            node = node.children.get(w)
            if not node:
                return False
        return node.isWord
    
class Solution(object):
    def findWords(self, board, words):
        """
        :type board: List[List[str]]
        :type words: List[str]
        :rtype: List[str]
        """
        res = []
        trie = Trie()
        node = trie.root
        for w in words:
            trie.insert(w)
        for i in range(len(board)):
            for j in range(len(board[0])):
                self.dfs(board, node, i, j, "", res)
        return sorted(res)
    
    def dfs(self, board, node, i, j, path, res):
        if node.isWord:
            res.append(path)
            node.isWord = False
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]):
            return 
        tmp = board[i][j]
        node = node.children.get(tmp)
        if not node:
            return 
        board[i][j] = "#"
        self.dfs(board, node, i+1, j, path+tmp, res)
        self.dfs(board, node, i-1, j, path+tmp, res)
        self.dfs(board, node, i, j-1, path+tmp, res)
        self.dfs(board, node, i, j+1, path+tmp, res)
        board[i][j] = tmp