79. Word Search

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。

Words must be in alphabetical order and formed by letters in adjacent cells, where "adjacent" cells are those that are adjacent horizontally or vertically. Letters in the same cell are not allowed to be reused.

 

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

 

prompt:

• boardAnd wordit contains only uppercase and lowercase letters.
• 1 <= board.length <= 200
• 1 <= board[i].length <= 200
• 1 <= word.length <= 10^3

DFS+backtracking

When you see a two-dimensional table, you think of DFS. It traverses the entire two-dimensional table. DFS is executed when the current position is the beginning letter of word. At the same time, a vis array is added to indicate the visited position, and then some judgment conditions are added.

Code

	def exist(self, board: List[List[str]], word: str) -> bool:
		def dfs(x, y, tmp):
			nonlocal flag
			if tmp == word:
				flag = True
				return
			if not (0 <= x < rowLength and 0 <= y < colLength):
				return
			if flag or vis[x][y] or board[x][y] != word[len(tmp)]:
				return
			vis[x][y] = True
			dfs(x + 1, y, tmp + board[x][y])
			dfs(x - 1, y, tmp + board[x][y])
			dfs(x, y + 1, tmp + board[x][y])
			dfs(x, y - 1, tmp + board[x][y])
			vis[x][y] = False

		rowLength, colLength, flag = len(board), len(board[0]), False
		vis = [[False for _ in range(colLength)] for _ in range(rowLength)]
		for i in range(rowLength):
			for j in range(colLength):
				if board[i][j] == word[0]:
					dfs(i, j, '')
				if flag:
					return flag
		return flag