Leetcode 79 Word Search Python 实现

最新推荐文章于 2025-08-18 23:17:42 发布

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原文链接：http://www.cnblogs.com/watch-fly/p/leetcode_79_watchfly.html

文章标签：

#python #数据结构与算法

本文详细解析了LeetCode 79题——单词搜索的算法实现，通过深度优先搜索(DFS)在二维网格中查找指定单词，介绍了如何避免重复访问同一格子，并提供了Python代码示例。

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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.

修改了几次  其中部分思路借鉴于他人代码

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        max_row = len(board)-1
        max_col = len(board[0])-1
        def backtrack(row, col ,s):
            if len(s) == 0:
                return True
            locs = [[row-1,col],[row+1,col],[row,col-1],[row,col+1]]
            #判断上下左右可用的点，只要有一个能够返回True则此次dfs可以返回True
            for loc in locs:
                if loc[0]<0 or loc[0]>max_row or loc[1]<0 or loc[1]>max_col or board[loc[0]][loc[1]] != s[0]:
                    continue
                #借鉴其他代码，把当前位置置为非字母字符，表示已经使用过
                tmp = board[loc[0]][loc[1]]
                board[loc[0]][loc[1]] = "$"
                if backtrack(loc[0], loc[1], s[1:]):
                    return True
                board[loc[0]][loc[1]] = tmp
            return False
        
        for r in range(len(board)):
            for c in range(len(board[0])):
                if board[r][c] == word[0]:
                    tmp = board[r][c]
                    board[r][c] = "$"
                    if backtrack(r,c,word[1:]):
                        return True
                    board[r][c] = tmp
        return False