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79_word_search.py
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79_word_search.py
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class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
def backtracking(row: int, col: int, index: int) -> bool:
if board[row][col] != word[index]:
return False
if index == len(word) - 1:
return True
else:
visited[row][col] = 1
ans = False
for direction in direction_arr:
new_row = row + direction[0]
new_col = col + direction[1]
if (0 <= new_row < m) and (0 <= new_col < n) and (visited[new_row][new_col] == 0):
if backtracking(new_row, new_col, index+1) == True:
ans = True
break
visited[row][col] = 0
return ans
first_letter_position = []
m = len(board)
n = len(board[0])
direction_arr = [[1, 0], [-1, 0], [0, 1], [0, -1]]
visited = [[0 for j in range(n)] for i in range(m)]
ans = False
for i in range(m):
for j in range(n):
if backtracking(i, j, 0) == True:
ans = True
break
return ans
'''
This is my personal record of solving Leetcode Problems.
If you have any questions, please discuss them in [Issues](https://github.com/mengxinayan/leetcode/issues).
Copyright (C) 2020-2022 mengxinayan
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program. If not, see <http://www.gnu.org/licenses/>.
'''