79 单词搜索

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

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:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

给定 word = "ABCCED", 返回 true
给定 word = "SEE", 返回 true
给定 word = "ABCB", 返回 false

解法

• 回溯

class Solution:
    orientation = [(0, -1), (-1, 0), (1, 0), (0, 1)]
    def exist(self, board, word):
        m = len(board)
        if m == 0: return False
        n = len(board[0])
        mark = [[False for _ in range(n)] for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if self.DFS(board, word, 0, i, j, mark, m, n):
                    return True
        return False
    def DFS(self, board, word, index, start1, start2, mark, m, n):
        if index == len(word)-1:
            return word[index] == board[start1][start2] 
        if board[start1][start2] == word[index]:
            mark[start1][start2] = True
            for k in self.orientation:
                start_x = start1+k[0]
                start_y = start2+k[1]
                if 0 <= start_x < m and 0 <= start_y < n:
                    if not mark[start_x][start_y] and self.DFS(
                        board, word, index+1, start_x, start_y, mark, m, n):
                        return True
            mark[start1][start2] = False
        return False