79-Word Search

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题意

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cells, where “adjacent” cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true

Example 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

思路

• DFS + flood fill
• 找一个字母，做标记 -> 上下左右继续找

class Solution(object):
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.dfs_search(board, i, j, word, 0):
                    return True
        return False
    
    def dfs_search(self, board, i, j, word, index):
        if index >= len(word): return True
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]): return False
        if board[i][j] ==  word[index] :
            index += 1
            copy = board[i][j]
            board[i][j] = '#'
            res = self.dfs_search(board, i-1, j, word, index) or self.dfs_search(board, i+1, j, word, index) or  self.dfs_search(board, i, j-1, word, index) or self.dfs_search(board, i, j+1, word, index)
            board[i][j] = copy
            return res
        
        return False