[leetcode] 79. Word Search

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 and word 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