79. Word Search

最新推荐文章于 2024-02-12 20:14:25 发布

原创最新推荐文章于 2024-02-12 20:14:25 发布 · 58 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

185 篇文章

订阅专栏

本文详细解析了LeetCode上单词搜索问题的解决方案，通过深度优先搜索算法在二维网格中查找给定单词是否存在，展示了递归函数的应用及状态标记数组的使用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

https://leetcode.com/problems/word-search/description/

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

class Solution(object):
    def check(self, arr, row, col, rest_word):
        if self.board[row][col] != rest_word[0] or not arr[row][col]:
            return False
        if len(rest_word) == 1:
            return True
        arr[row][col] = False
        # print self.board[row][col], row, col, arr
        if row > 0 and self.check(arr, row - 1, col, rest_word[1:]):
            return True
        if col > 0 and self.check(arr, row, col - 1, rest_word[1:]):
            return True
        if row < len(self.board) - 1 and self.check(arr, row + 1, col, rest_word[1:]):
            return True
        if col < len(self.board[0]) - 1 and self.check(arr, row, col + 1, rest_word[1:]):
            return True
        arr[row][col] = True
        return False
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        arr = [[True] * len(board[0]) for _ in range(len(board))]
        self.board = board
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.check(arr, i, j, word):
                    return True
        return False