描述

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.

Python

很好理解的backtracking

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        if not board:return False
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.dfs(board,i,j,word):
                    return True
        return False
    
    # check whether can find word, start at (i,j) position
    def dfs(self,board,i,j,word):
        if len(word)==0:
            return True
        if i<0 or i>len(board)-1 or j<0 or j>len(board[0])-1 or word[0]!=board[i][j]:
            return False
        tmp = board[i][j] # first character is found, check the remaining part
        board[i][j] = '#' # avoid visit agian
        # check whether can find "word" along one direction
        res = self.dfs(board,i-1,j,word[1:]) \
        or self.dfs(board,i+1,j,word[1:]) \
        or self.dfs(board,i,j-1,word[1:]) \
        or self.dfs(board,i,j+1,word[1:])
        board[i][j] = tmp # do not change the origin matrix
        return res