LeetCode | 0079. Word Search單詞搜索【Python】

LeetCode 0079. Word Search單詞搜索【Medium】【Python】【DFS】

Problem

LeetCode

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 cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example:

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

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

問題

力扣

給定一個二維網格和一個單詞，找出該單詞是否存在於網格中。

單詞必須按照字母順序，通過相鄰的單元格內的字母構成，其中“相鄰”單元格是那些水平相鄰或垂直相鄰的單元格。同一個單元格內的字母不允許被重複使用。

示例:

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

給定 word = "ABCCED", 返回 true.
給定 word = "SEE", 返回 true.
給定 word = "ABCB", 返回 false.

思路

DFS

DFS 四個方向搜索，訪問過的標記爲 '#' 表示不可重複訪問。
記得訪問結束要恢復，因爲還有其他路徑。

Python3代碼

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        if not board:
            return False
        
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.dfs(board, i, j, word):
                    return True
        return False

    def dfs(self, board, i, j, word):
        if len(word) == 0:
            return True
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or word[0] != board[i][j]:
            return False
        tmp, board[i][j] = board[i][j], '#'  # '#': visited
        ans = self.dfs(board, i - 1, j, word[1:]) or self.dfs(board, i + 1, j, word[1:]) or self.dfs(board, i, j - 1, word[1:]) or self.dfs(board, i, j + 1, word[1:])
        board[i][j] = tmp  # recover board[i][j]
        return ans

代碼地址

GitHub鏈接