[leetcode]79. Word Search

79. Word Search

 

Given a 2D board and a word, find if the word exists in the grid.

The word can 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.

Example:

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

Given word = "ABCCED", return true.

Given word = "SEE", return true.

Given word = "ABCB", return false.

 

class Solution(object):
    def dfs(self, board, word, point):
        if not word:
            return True

        direct = [[-1,0,1,0],[0,-1,0,1]]
        for i in range(4):
            cur_y = point[0]+ direct[0][i]
            cur_x = point[1]+ direct[1][i]
            if cur_y >=0 and cur_y < len(board) and cur_x >=0 and cur_x < len(board[0]) and board[cur_y][cur_x]== word[0]:
                origin = board[cur_y][cur_x]
                board[cur_y][cur_x] = '0'
                result = self.dfs(board, word[1:], (cur_y, cur_x))
                if result : 
                    return True
                board[cur_y][cur_x] = origin 
        return False

    def exist(self, board, word):
        if len(board) == 0 or len(board[0])== 0:
            return False
        
        for i in range(len(board)):
            for j in range(len(board[0])):
                if board[i][j] == word[0]:
                    origin = board[i][j]
                    board[i][j] = '0'
                    result = self.dfs(board, word[1:], (i,j))
                    if result:
                        return True
                    board[i][j] = origin 
        return False
        

https://leetcode.com/problems/word-search/