#79 Word Search——Top 100 Liked Questions

最新推荐文章于 2025-01-20 14:51:04 发布

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最新推荐文章于 2025-01-20 14:51:04 发布

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分类专栏： LeetCode

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博客讨论在二维网格中查找单词是否存在的问题。单词需由相邻单元格字母构成，且同一单元格字母不能重复使用。采用深度优先搜索递归（dfs）方法实现，给出了Python代码的运行时间和内存使用情况。

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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.

"""

第一次：深度优先搜索递归dfs
"""

class Solution(object):
    def dfs(self, board, i, j, word):

        if len(word) == 0:
            return True #全部查找到
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or board[i][j] != word[0]:
            return False
        tmp = board[i][j]
        board[i][j] = '#' #为了避免重复查找，用井号代替
        res = self.dfs(board, i - 1, j, word[1: ]) or self.dfs(board, i + 1, j, word[1: ]) or self.dfs(board, i, j - 1, word[1: ]) or self.dfs(board, i, j + 1, word[1: ])
        board[i][j] = tmp
        return res
    
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        if not board:
            return False
        for i in range(len(board)):
            for j in range(len(board[0])):#从任意位置出发进行深度搜索
                if self.dfs(board, i, j, word):
                    return True  #如果能找到，返回true
        return False

"""

Runtime: 344 ms, faster than 52.69% of Python online submissions forWord Search.

Memory Usage: 16.3 MB, less than 29.48% of Python online submissions for Word Search.

"""