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[11/5] 212. Word Search II

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작성일 2022.11.05 10:46

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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.

Example 1:

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"]

Example 2:

Input: board = [["a","b"],["c","d"]], words = ["abcb"]
Output: []

Constraints:

m == board.length
n == board[i].length
1 <= m, n <= 12
board[i][j] is a lowercase English letter.
1 <= words.length <= 3 * 104
1 <= words[i].length <= 10
words[i] consists of lowercase English letters.
All the strings of words are unique.

Accepted

512,675

Submissions

1,388,063

#Amazon, #Uber, #Cisco, #Microsoft, #Bloomberg, #Apple, #Karat, #Facebook, #Google, #Indeed, #TikTok

링크

https://leetcode.com/problems/word-search-ii/ 1248 회 연결

익명
작성일 2022.11.05 10:46

class TrieNode:
    def __init__(self):
        self.word = None
        self.children = {}
        
class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trieroot = TrieNode()
        
        for word in words:
            node = trieroot
            
            for idx, char in enumerate(word):
                if char not in node.children:
                    node.children[char] = TrieNode()
                node = node.children[char]
                if idx == len(word)-1:
                    node.word = word
        
        directions = [(1,0), (0,1), (-1,0), (0,-1)]
        
        ROW, COL = len(board), len(board[0])
        path = set()
        
        res = []
        
        def dfs(row, col, trienode):
            if trienode.word:
                res.append(trienode.word)
                trienode.word = None
            path.add((row,col))
            
            for rd, cd in directions:
                nr, nc = row+rd, col+cd
                if 0<=nr<ROW and 0<=nc<COL and (nr,nc) not in path and board[nr][nc] in trienode.children:
                    dfs(nr,nc,trienode.children[board[nr][nc]])
            
            path.remove((row,col))
        
        for r in range(ROW):
            for c in range(COL):
                if board[r][c] in trieroot.children:
                    node = trieroot
                    dfs(r,c,node.children[board[r][c]])
        return res

[code=Python 3]class TrieNode:
    def __init__(self):
        self.word = None
        self.children = {}
        
class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trieroot = TrieNode()
        
        for word in words:
            node = trieroot
            
            for idx, char in enumerate(word):
                if char not in node.children:
                    node.children[char] = TrieNode()
                node = node.children[char]
                if idx == len(word)-1:
                    node.word = word
        
        directions = [(1,0), (0,1), (-1,0), (0,-1)]
        
        ROW, COL = len(board), len(board[0])
        path = set()
        
        res = []
        
        def dfs(row, col, trienode):
            if trienode.word:
                res.append(trienode.word)
                trienode.word = None
            path.add((row,col))
            
            for rd, cd in directions:
                nr, nc = row+rd, col+cd
                if 0<=nr<ROW and 0<=nc<COL and (nr,nc) not in path and board[nr][nc] in trienode.children:
                    dfs(nr,nc,trienode.children[board[nr][nc]])
            
            path.remove((row,col))
        
        for r in range(ROW):
            for c in range(COL):
                if board[r][c] in trieroot.children:
                    node = trieroot
                    dfs(r,c,node.children[board[r][c]])
        return res[/code]