1. 题目
2. 分析
这题其实非常不错。如果正向解，非常麻烦；因为很难界定哪些O是被包围的？但是如果反向解呢？因为边界的O不会被包围，那么就可以想到跟边界O相连的O都不会被包围。那么除此之外的O都会被包围，题目就解决了。
3. 代码
class Solution:
def solve(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
m, n = len(board), len(board[0])
vis = [[0] * n for i in range(m)]
# 只从边界遍历
for i in [0,m-1]:
for j in range(n):
if board[i][j] == 'O':
self.dfs(i, j, m, n, vis, board)
for j in [0, n-1]:
for i in range(m):
if board[i][j] == 'O':
self.dfs(i, j, m, n, vis, board)
print(vis)
for i in range(m):
for j in range(n):
if vis[i][j] == 0:
board[i][j] = 'X'
def dfs(self, i, j, m, n, vis, board):
if i>=0 and j>=0 and i<m and j< n:
if vis[i][j] == 0 and board[i][j] == "O":
vis[i][j] = 1
for item in [(i-1,j), (i, j-1), (i+1, j), (i, j+1)]:
new_i, new_j = item
self.dfs(new_i, new_j, m, n, vis, board)
