LeetCode----Word Search-优快云博客

本文链接：https://blog.youkuaiyun.com/whiterbear/article/details/49001765

本文详细介绍了二维矩阵中寻找指定单词的算法实现，包括矩阵构造、DFS遍历及优化策略，旨在解决如何在给定矩阵内找到特定单词的问题。

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Word Search

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.

For example,
Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

分析：

这题题目意思描述的不对。

当borad为[ ["ABCE"], ["SFCS"], ["ADEE"] ] 时，其应该为[ ["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"] ] 。

本题使用DFS来做。

代码：

class Solution(object):
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        lines = len(board)
        rows = len(board[0])
        # 先判断board中的字符数是否足够word来访问
        if rows * lines < len(word):
            return False

        # 记录word中第一个字符在board中的位置，将从这些位置开始dfs遍历
        existposlst = []
        for i, l in enumerate(board):
            for j, r in enumerate(l):
                if r == word[0]:
                    existposlst.append([i, j])

        # 开始DFS
        for pos in existposlst:
            if self.dfs(board, rows, lines, word[1:], visistposlst=[pos], curpos=pos):
                return True
        return False

    def dfs(self, board, rows, lines, word, visistposlst, curpos):
        """
        :visistposlst: 列表类型，记录当前已经访问过的位置，不可重复访问（注意：集合类型无法hashable list）
        :curpos: 如[0, 1], 保存当前访问的位置
        :rtype: bool
        """
        if word == '':
            return True
        movedirect = [[-1, 0], [1, 0], [0, -1], [0, 1]]  # 上下左右
        for direct in movedirect:
            newpos = [curpos[0] + direct[0], curpos[1] + direct[1]]
            if 0 <= newpos[0] < lines and 0 <= newpos[1] < rows and board[newpos[0]][newpos[1]] == word[0]:
                if newpos not in visistposlst:
                    if self.dfs(board, rows, lines, word[1:], visistposlst + [newpos], newpos):
                        return True
        return False

后记：

本题我的代码效率并不高，但是却又找不到优化的办法。