[Leetcode]@python 79. Word Search

题目链接

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

题目原文

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 =

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

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

题目大意

给出一字表和词语，返回这个词语是否在字表上。字表之间只能水平和垂直相连，字表中的字符只能访问一次

解题思路

使用dfs来搜索，为了避免已经用到的字母被重复搜索，将已经用到的字母临时替换为'#'

代码

class Solution(object):
    def exist(self, board, word):
        """
 :type board: List[List[str]]
 :type word: str
 :rtype: bool
 """

        def dfs(x, y, word):
            if len(word) == 0:
                return True
            # up
            if x > 0 and board[x - 1][y] == word[0]:
                tmp = board[x][y]
                board[x][y] = '#'
                if dfs(x - 1, y, word[1:]):
                    return True
                board[x][y] = tmp

            # down
            if x < len(board) - 1 and board[x + 1][y] == word[0]:
                tmp = board[x][y]
                board[x][y] = '#'
                if dfs(x + 1, y, word[1:]):
                    return True
                board[x][y] = tmp

            # left
            if y > 0 and board[x][y - 1] == word[0]:
                tmp = board[x][y]
                board[x][y] = '#'
                if dfs(x, y - 1, word[1:]):
                    return True
                board[x][y] = tmp

            # right
            if y < len(board[0]) - 1 and board[x][y + 1] == word[0]:
                tmp = board[x][y]
                board[x][y] = '#'
                if dfs(x, y + 1, word[1:]):
                    return True
                board[x][y] = tmp

            return False

        for i in range(len(board)):
            for j in range(len(board[0])):
                if board[i][j] == word[0]:
                    if (dfs(i, j, word[1:])):
                        return True
        return False