[LeetCode]79 单词搜索

题目描述

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.

代码

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class Solution:
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        def check(board, word):
            d = collections.defaultdict(int)
            m, n = len(board), len(board[0])
            for i in range(m):
                for j in range(n):
                    d[board[i][j]] += 1
            for c in word:
                if c not in d or d[c] <= 0:
                    return False
                d[c] -= 1
            return True
        
        def dfs(board, i, j, word):
            c = board[i][j]
            board[i][j] = '*'
            m, n = len(board), len(board[0])
            if not word:
                return True
            for di, dj in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                if i+di < 0 or i+di >= m or j+dj < 0 or j+dj >= n:
                    continue
                if board[i+di][j+dj] == word[0] and dfs(board, i+di, j+dj, word[1:]):
                    return True
            board[i][j] = c
            return False
        if not word:
            return True
        if not board or not board[0]:
            return False
        if not check(board, word):
            return False
        last = word[1:]
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                if board[i][j] == word[0] and dfs(board, i, j, last):
                    return True
        return False