79. 单词搜索

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？

Python 解答：
1.回溯

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        m = len(board)
        n = len(board[0])
        lis = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        flag = [[True for i in range(n)] for j in range(m)]
        def check(board, i, j, word, k, temp):
            if board[i][j] != word[k]:
                return False
            else:
                temp += board[i][j]
                flag[i][j] = False
                if k == len(word)-1:
                    return True
                for item in lis:
                    newi = i + item[0]
                    newj = j + item[1]
                    if 0 <= newi < m and 0 <= newj < n and flag[newi][newj]:
                        if check(board, newi, newj, word, k+1, temp):
                            return True
                flag[i][j] = True
                temp = temp[:-1]
                return False
        for i in range(m):
            for j in range(n):
                if check(board, i, j, word, 0, ""):
                    return True
        return False