API 介面應該如何設計?如何保證安全?如何簽名?如何防重?
阿新 • • 發佈:2022-03-29
題目
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,搜尋路徑
程式碼
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