【leetcode系列】【算法】【中等】单词搜索

题目：

题目链接： https://leetcode-cn.com/problems/word-search/

解题思路：

先对二维数组遍历寻找单词的首个字母，如果找到，递归的对当前位置的上下左右查找单词的后续字母

图解如下(假设搜索方向顺序为上 → 下 → 左 → 右)：

代码实现：

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def __search(board, row, column, word, start):
            if start >= len(word):
                return True
            
            # up
            if row > 0 and board[row - 1][column] == word[start]:
                board[row - 1][column] = '0'
                if True == __search(board, row - 1, column, word, start + 1):
                    return True
                
                board[row - 1][column] = word[start]
                
            # down
            if row < len(board) - 1 and board[row + 1][column] == word[start]:
                board[row + 1][column] = '0'
                if True == __search(board, row + 1, column, word, start + 1):
                    return True
                
                board[row + 1][column] = word[start]

            # left
            if column > 0 and board[row][column - 1] == word[start]:
                board[row][column - 1] = '0'
                if True == __search(board, row, column - 1, word, start + 1):
                    return True
                
                board[row][column - 1] = word[start]
                
            # right
            if column < len(board[0]) - 1 and board[row][column + 1] == word[start]:
                board[row][column + 1] = '0'
                if True == __search(board, row, column + 1, word, start + 1):
                    return True
                
                board[row][column + 1] = word[start]

            return False

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