剑指 Offer 12. 矩阵中的路径

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

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","d"]], word = "abcd"
输出：false

提示：

• 1 <= board.length <= 200
• 1 <= board[i].length <= 200

Python 解答：

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        length = len(board)
        width = len(board[0])
        flag = [[0 for i in range(width)] for j in range(length)]
        index = 0
        temp = [[i,j] for i in range(length) for j in range(width)]
        def dfs(board, word, flag, points, index):
            if index == len(word):
                return True
            res = False
            for point in points:
                temp = []
                if board[point[0]][point[1]] == word[index]:
                    flag[point[0]][point[1]] = 1
                    if point[0]+1 < len(board) and flag[point[0]+1][point[1]] == 0:
                        temp.append([point[0]+1, point[1]])
                    if point[0]-1 >= 0 and flag[point[0]-1][point[1]] == 0:
                        temp.append([point[0]-1, point[1]])
                    if point[1]+1 < len(board[0]) and flag[point[0]][point[1]+1] == 0:
                        temp.append([point[0], point[1]+1])
                    if point[1]-1 >= 0 and flag[point[0]][point[1]-1] == 0:
                        temp.append([point[0], point[1]-1])
                    res = res or dfs(board, word, flag, temp, index+1)
                    flag[point[0]][point[1]] = 0
            return res
        return dfs(board, word, flag, temp, 0)