12_StringPathInMatrix.py

# -*- coding: utf-8 -*-
# @File    : 12_StringPathInMatrix.py
# @Date    : 2020-10-16
# @Author  : tc
"""
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
注意：本题与主站 79 题相同：https://leetcode-cn.com/problems/word-search/

参考：https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/solution/mian-shi-ti-12-ju-zhen-zhong-de-lu-jing-shen-du-yo/

"""
from typing import List

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def dfs(i,j,k):
            if not 0 <= i < len(board) or not 0 <= j <= len(board[0]) or board[i][j] != word[k]:
                return False
            if k == len(word) - 1:
                return True
            tmp, board[i][j] = board[i][j], '/'
            res = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)
            board[i][j] = tmp
            return res

        for i in range(len(board)):
            for j in range(len(board[0])):
                if dfs(i, j, 0):
                    return True
        return False

if __name__ == '__main__':
    pass