LeetCode题解：Surrounded Regions

Given a 2D board containing ‘X’ and ‘O’, capture all regions surrounded by ‘X’.

A region is captured by flipping all ‘O’s into ‘X’s in that surrounded region.

For 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

题意：给定由X和O组成的二维数组，将所有被X包围的O替换成X

解决思路：BFS和DFS都可以

代码：

public class Solution {
private int row = 0;
private int col = 0;
private Queue<Integer> queue = new LinkedList<Integer>();

public void solve(char[][] board) {
if(board == null || board.length == 0){
return;
}

row = board.length;
col = board[0].length;
boolean[][] mark = new boolean[row][col];

for(int i = 0; i < row; i++){
bfs(i, 0, board, mark);
bfs(i, col - 1, board, mark);
}

for(int i = 0; i < col; i++){
bfs(0, i, board, mark);
bfs(row - 1, i, board, mark);
}

for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
if(board[i][j] == 'O'){
board[i][j] = 'X';
}else if(board[i][j] == 'C'){
board[i][j] = 'O';
}
}
}
}

private void bfs(int x, int y, char[][] board, boolean[][] mark){
mark(x, y, board, mark);

while(!queue.isEmpty()){
int curr = queue.poll();
int xPos = curr / col;
int yPos = curr % col;

mark(xPos + 1, yPos, board, mark);
mark(xPos, yPos + 1, board, mark);
mark(xPos - 1, yPos, board, mark);
mark(xPos, yPos - 1, board, mark);
}
}

private void mark(int x, int y, char[][] board, boolean[][] mark){
if(x < 0 || y < 0 || x >= row || y >= col || mark[x][y] == true || board[x][y] != 'O'){
return;
}

mark[x][y] = true;
board[x][y] = 'C';
queue.offer(x * col + y);
}
}