[Python知识库]剑指offer-python：10.矩阵中单词的路径

题目：

给定一个M*N的字符网格-board，一个字符串word，如果word路径在网格中存在，返回true。

例如：

1.输入： board = [["A","B","C","E"],["S","F","C","S"],["A","D", "E","E"]], word = "ABCCED"；

? ?输出：true

????????

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

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

约束条件：

1.m == board.length ；n = board[i].length

2.1 <= m, n <= 6

3.1 <= word.length <= 15

4.board并且word仅由小写和大写英文字母组成。?

解析：可参考?

https://darktiantian.github.io/%E7%9F%A9%E9%98%B5%E4%B8%AD%E5%8D%95%E8%AF%8D%E7%9A%84%E8%B7%AF%E5%BE%84%EF%BC%8C%E5%BE%88%E5%A4%9A%E4%BA%BA%E9%83%BD%E9%94%99%E4%BA%86/

代码：

def find_word(board , word):
    m , n = len(board) , len(board[0])

    def find_other(i, j, word_other):
        if not word_other:
            return True
        mid_flag = False
        origional, board[i][j] = board[i][j], '-'
        for new_i, new_j in ((i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)):
            if 0 <= new_i < m and 0 <= new_j < n:
                if word_other[0] == board[new_i][new_j]:
                    if find_other(new_i, new_j,  word_other[1:]):
                        mid_flag = True
                        break
        board[i][j] = origional
        return mid_flag

    for mi in range(m):
        for ni in range(n):
            if board[mi][ni] == word[0]:
                if find_other(mi , ni , word[1:] ):
                    return True
    return False

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