题目
Given an m x n
grid of characters board
and a string word
, return true
if word
exists in the grid.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true
Example 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true
Example 3:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false
Constraints:
m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board
andword
consists of only lowercase and uppercase English letters.
Follow up: Could you use search pruning to make your solution faster with a larger board
?
思路
dfs,搜索路径
代码
python版本:
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
used = [[0 for _ in range(len(board[0]))] for _ in range(len(board))]
res = False
def dfs(i, j, n):
nonlocal res
if n >= len(word):
res = True
return
if res or i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or used[i][j]:
return
if board[i][j] != word[n]:
return
for x, y in ((0, 1), (0, -1), (1, 0), (-1, 0)):
used[i][j] = 1
dfs(i+x, j+y, n+1)
used[i][j] = 0
[dfs(i, j, 0) for i in range(len(board)) for j in range(len(board[0]))]
return res