题目
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.
解题思路
遍历整个board，如果和要找的单词的第一个字母相同，就开始递归回溯
代码
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
def dfs(x, y, p):
if p == l:
return True
for dirx, diry in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
curx, cury = x + dirx, y + diry
if 0 <= curx < m and 0 <= cury < n and flag[curx][cury] and board[curx][cury] == word[p]:
flag[curx][cury] = False
if dfs(curx, cury, p + 1):
flag[curx][cury] = True
return True
flag[curx][cury] = True
return False
m, n = len(board), len(board[0])
l = len(word)
flag = [[True] * n for _ in range(m)]
for i in range(m):
for j in range(n):
if board[i][j] == word[0]:
flag[i][j] = False
if dfs(i, j, 1):
return True
flag[i][j] = True
return False
