Problem description
Given a 2D board containing ‘X’ and ‘O’ (the letter O), capture all regions surrounded by ‘X’.
A region is captured by flipping all ‘O’s into ‘X’s in that surrounded region.
Example:
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:
X X X X
X X X X
X X X X
X O X X
Explanation:
Surrounded regions shouldn’t 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.
Solution
We can use DFS to mark all the ‘O’ s that connected to the boarder and change the non-marked ‘O’s to ‘X’
Below is my python implementation
class Solution:
def solve(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
m = len(board)
if m == 0:
return
n = len(board[0])
def neighbors(x, y)->List[List[int]]:
ans = []
if x + 1 < m and board[x + 1][y] == 'O':
ans.append([x + 1, y])
if x - 1 >= 0 and board[x - 1][y] == 'O':
ans.append([x - 1, y])
if y + 1 < n and board[x][y + 1] == 'O':
ans.append([x, y + 1])
if y - 1 >= 0 and board[x][y - 1] == 'O':
ans.append([x, y - 1])
return ans
def dfs(i:int, j:int)->None: #mark O as S then changed it back to O
board[i][j] = 'S'
for neighbor in neighbors(i,j):
x, y = neighbor
dfs(x, y)
return
for i in range(m):
if board[i][0] == 'O':
dfs(i, 0)
if board[i][n - 1] == 'O':
dfs(i, n - 1)
for j in range(n):
if board[0][j] == 'O':
dfs(0, j)
if board[m - 1][j] == 'O':
dfs(m - 1, j)
for i in range(m):
for j in range(n):
if board[i][j] == 'O':
board[i][j] = 'X'
if board[i][j] == 'S':
board[i][j] = 'O'
return