题目描述

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

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

解答方法

方法一：dfs回溯法

思路

https://leetcode-cn.com/problems/word-search/solution/zai-er-wei-ping-mian-shang-shi-yong-hui-su-fa-pyth/

代码

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        directs = [[-1,0],[1,0],[0,-1],[0,1]]
        m = len(board)
        if m == 0:
            return False
        n = len(board[0])
        mark = [[False for _ in range(n)] for _ in range(m)]
        def backtrack(x, y,index):
            if index == len(word) -1:
                return board[x][y] == word[index]
            if board[x][y] == word[index]:
                mark[x][y] = True
                for direct in directs:
                    cur_x = x + direct[0]
                    cur_y = y + direct[1]
                    if cur_x >=0 and cur_x < m and cur_y >= 0 and cur_y < n and not mark[cur_x][cur_y] and backtrack( cur_x, cur_y,  index+1):
                        return True
                mark[x][y] = False
            return False

        for i in range(m):
            for j in range(n):
                if backtrack( i, j, 0):
                    return True
        return False

时间复杂度

O((m*n)^2)

空间复杂度

O(m*n)