Question
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.
Solution
class Solution:
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
if len(word) == 0:
return True
if len(board) == 0 or len(board[0]) == 0:
return False
x, y = len(board), len(board[0])
char = word[0]
for i in range(x):
for j in range(y):
if board[i][j] == char:
ret = self.backtracking_one_step(board, x, y, word, 0, i, j, char)
if ret:
return True
return False
def backtracking(self, board, x, y, word, index, i, j):
if index == len(word) - 1:
return True
index += 1
char = word[index]
if i > 0 and board[i - 1][j] == char:
if self.backtracking_one_step(board, x, y, word, index, i - 1, j, char):
return True
if i < x - 1 and board[i + 1][j] == char:
if self.backtracking_one_step(board, x, y, word, index, i + 1, j, char):
return True
if j > 0 and board[i][j - 1] == char:
if self.backtracking_one_step(board, x, y, word, index, i, j - 1, char):
return True
if j < y - 1 and board[i][j + 1] == char:
if self.backtracking_one_step(board, x, y, word, index, i, j + 1, char):
return True
return False
def backtracking_one_step(self, board, x, y, word, index, i, j, char):
board[i][j] = ''
ret = self.backtracking(board, x, y, word, index, i, j)
if ret:
return True
else:
board[i][j] = char
return False