79.py

'''
 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.
'''

classSolution(object):
defexist(self, board, word):
"""
 :type board: List[List[str]]
 :type word: str
 :rtype: bool
 """

result=False
forrowinrange(len(board)):
forcolinrange(len(board[0])):
ifself.dfs(board, word, row, col, 0):
returnTrue
returnFalse

defdfs(self, board, word, row, col, curr_len):
ifrow<0orcol<0orrow>=len(board) orcol>=len(board[0]):
returnFalse
ifboard[row][col] ==word[curr_len]:
c=board[row][col]
board[row][col] ='#'

ifcurr_len==len(word) -1:
returnTrue
elif (self.dfs(board, word, row-1, col, curr_len+1) orself.dfs(board, word, row+1, col, curr_len+1) orself.dfs(board, word, row, col-1, curr_len+1) orself.dfs(board, word, row, col+1, curr_len+1)):
returnTrue

board[row][col] =c
returnFalse