https://leetcode.com/problems/word-search/
Medium
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.
- 经典回溯。注意Python列表初始化，复习DFS模版。
1 class Solution: 2 def exist(self, board: List[List[str]], word: str) -> bool: 3 if not word: 4 return False 5 6 visited = [ [ False for j in range( len( board[ 0 ] ) ) ] for i in range( len( board ) ) ] # remember the way 7 8 def helper(x, y, current): 9 if current == len( word ): 10 return True 11 12 if x < 0 or x >= len( board ) or y < 0 or y >= len( board[ 0 ] ) or visited[x][y] or board[x][y] != word[current]: 13 return False 14 15 visited[x][y] = True 16 17 result = (helper(x - 1, y, current + 1) 18 or helper(x + 1, y, current + 1) 19 or helper(x, y - 1, current + 1) 20 or helper(x, y + 1, current + 1) ) 21 22 visited[x][y] = False 23 24 return result 25 26 for i in range( len( board ) ): 27 for j in range( len ( board[ 0 ] ) ): 28 if helper(i, j, 0): 29 return True 30 31 return False