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.
修改了几次 其中部分思路借鉴于他人代码
class Solution: def exist(self, board: List[List[str]], word: str) -> bool: max_row = len(board)-1 max_col = len(board[0])-1 def backtrack(row, col ,s): if len(s) == 0: return True locs = [[row-1,col],[row+1,col],[row,col-1],[row,col+1]] #判断上下左右可用的点，只要有一个能够返回True则此次dfs可以返回True for loc in locs: if loc[0]<0 or loc[0]>max_row or loc[1]<0 or loc[1]>max_col or board[loc[0]][loc[1]] != s[0]: continue #借鉴其他代码，把当前位置置为非字母字符，表示已经使用过 tmp = board[loc[0]][loc[1]] board[loc[0]][loc[1]] = "$" if backtrack(loc[0], loc[1], s[1:]): return True board[loc[0]][loc[1]] = tmp return False for r in range(len(board)): for c in range(len(board[0])): if board[r][c] == word[0]: tmp = board[r][c] board[r][c] = "$" if backtrack(r,c,word[1:]): return True board[r][c] = tmp return False
Leetcode 79 Word Search Python 实现
原文：https://www.cnblogs.com/watch-fly/p/leetcode_79_watchfly.html