Given an m x n matrix board containing ‘X’ and ‘O’, capture all regions that are 4-directionally surrounded by ‘X’. A region is captured by flipping all ‘O’s into ‘X’s in that surrounded region.
Example 1:
Input: board = [[“X”,“X”,“X”,“X”],[“X”,“O”,“O”,“X”],[“X”,“X”,“O”,“X”],[“X”,“O”,“X”,“X”]] Output: [[“X”,“X”,“X”,“X”],[“X”,“X”,“X”,“X”],[“X”,“X”,“X”,“X”],[“X”,“O”,“X”,“X”]] Explanation: Surrounded regions should not be on the border, which means that any ‘O’ on the border of the board are not flipped to ‘X’. Any ‘O’ that is not on the border and it is not connected to an ‘O’ on the border will be flipped to ‘X’. Two cells are connected if they are adjacent cells connected horizontally or vertically.
Example 2:
Input: board = [[“X”]] Output: [[“X”]]
Constraints:
m == board.length
n == board[i].length
1 <= m, n <= 200
board[i][j] is 'X' or 'O'.
- code
class Solution:
def solve(self, board: List[List[str]]) -> None:
m = len(board)
n = len(board[0])
memo = set()
def dfs(x, y):
if board[x][y] == "X": return
memo.add((x, y))
for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < m and 0 <= ny < n and board[nx][ny] == "O" and (nx, ny) not in memo:
dfs(nx, ny)
for i in range(m):
dfs(i, 0)
dfs(i, n - 1)
for j in range(n):
dfs(0, j)
dfs(m - 1, j)
for i, j in product(range(m), range(n)):
if board[i][j] == "O" and (i, j) not in memo:
board[i][j] = "X"
- code use a mark, instead of memo
class Solution:
def solve(self, board: List[List[str]]) -> None:
m = len(board)
n = len(board[0])
def dfs(x, y):
if board[x][y] == "X": return
board[x][y] = "R" # change O to R
for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < m and 0 <= ny < n and board[nx][ny] == "O":
dfs(nx, ny)
for i in range(m):
dfs(i, 0)
dfs(i, n - 1)
for j in range(n):
dfs(0, j)
dfs(m - 1, j)
for i, j in product(range(m), range(n)):
if board[i][j] == "O": board[i][j] = "X"
elif board[i][j] == "R": board[i][j] = "O"