Problem
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 1
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X X X X
X O O X
X X O X
X O X X
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After running your function, the board should be:
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X X X X
X X X X
X X X X
X O X X
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Solution
DFS
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func solve(board [][]byte) {
a := make([][]int, len(board))
for idx := range board {
a[idx] = make([]int, len(board[idx]))
}
for idx := range board {
if idx == 0 || idx == len(board)-1 {
for jdx := range board[idx] {
if board[idx][jdx] == 'O' && a[idx][jdx] == 0 {
visit(board, a, idx, jdx)
}
}
} else {
jdx := 0
if board[idx][jdx] == 'O' && a[idx][jdx] == 0 {
visit(board, a, idx, jdx)
}
jdx = len(board[idx]) - 1
if board[idx][jdx] == 'O' && a[idx][jdx] == 0 {
visit(board, a, idx, jdx)
}
}
}
for idx := range board {
for jdx := range board[idx] {
if a[idx][jdx] != 1 {
board[idx][jdx] = 'X'
}
}
}
}
func visit(board [][]byte, a [][]int, i, j int) {
a[i][j] = 1
if j+1 < len(board[i]) && board[i][j+1] == 'O' && a[i][j+1] == 0 {
visit(board, a, i, j+1)
}
if j-1 >= 0 && board[i][j-1] == 'O' && a[i][j-1] == 0 {
visit(board, a, i, j-1)
}
if i+1 < len(board) && board[i+1][j] == 'O' && a[i+1][j] == 0 {
visit(board, a, i+1, j)
}
if i-1 >= 0 && board[i-1][j] == 'O' && a[i-1][j] == 0 {
visit(board, a, i-1, j)
}
}
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