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word_search.py
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word_search.py
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"""
79. 单词搜索
数组 回溯 矩阵
中等
给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中,返回 true ;否则,返回 false 。
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 1:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出:true
示例 2:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出:true
示例 3:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出:false
提示:
m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board 和 word 仅由大小写英文字母组成
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/word-search
"""
from typing import List
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
def dfs(i: int, j: int, k: int):
if board[i][j] != word[k]:
return False
if k == size - 1:
return True
visited.add((i, j))
res = False
for di, dj in ((0, -1), (0, 1), (-1, 0), (1, 0)):
ni, nj = i+di, j+dj
if 0 <= ni < m and 0 <= nj < n:
if (ni, nj) not in visited:
if dfs(ni, nj, k+1):
res = True
break
visited.remove((i, j))
return res
m, n = len(board), len(board[0])
size = len(word)
visited = set()
for x in range(m):
for y in range(n):
if dfs(x, y, 0):
return True
return False
if __name__ == '__main__':
solution = Solution()
result = solution.exist([["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], "ABCCED")
print(result)
assert result is True
result = solution.exist([["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], "SEE")
print(result)
assert result is True
result = solution.exist([["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], "ABCB")
print(result)
assert result is False