[LeetCode]题解（python）：079 Word Search

题目来源

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https://leetcode.com/problems/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.

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题意分析

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Input:

　　:type board: List[List[str]]
　　:type word: str

Output:

:rtype: bool

Conditions:给出一个字符串，看这个字符串是否可以在board中找到，找到的条件是：在board可以连成一条直线，注意已经用过的元素不能再用。

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题目思路

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用dfs，每次遍历上下左右（注意边界）。避免使用过的元素再次使用，则可以将已经使用过的元素替换为“#”，在使用完之后再替换回来。

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AC代码（Python）

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 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

             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

             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

             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

             if y < len(board[x]) - 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[i])):
                 if board[i][j] == word[0]:
                     if dfs(i, j, word[1:]):
                         return True
         return False