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

核心的思想就是边上的O都不能改变，深度搜索与这样的点相邻的点，保持不变。

public void solve(char[][] board) {
if(board == null || board.length==0)
return;
int len1=board.length,len2=board[0].length;
for(int i=0;i<len1;i++){
if(board[i][0]=='O')
merge(board,i,0);
if(board[i][len2-1]=='O')
merge(board,i,len2-1);
}
for(int i=0;i<len2;i++){
if(board[0][i]=='O')
merge(board,0,i);
if(board[len1-1][i]=='O')
merge(board,len1-1,i);
}
for(int i=0;i<len1;i++){
for(int j=0;j<len2;j++){
if(board[i][j]=='O')
board[i][j]='X';
else if(board[i][j]=='*')
board[i][j]='O';
}
}
}
public void merge(char[][] board, int i, int j){
if(i<0 || i>=board.length || j<0 || j>=board[0].length)
return;
if(board[i][j] != 'O')
return;
board[i][j] = '*';
merge(board, i-1, j);
merge(board, i+1, j);
merge(board, i, j-1);
merge(board, i, j+1);
}

上面的代码当board很大时，就会出现java.lang.StackOverflowError的错误，所有

改用一个队列来存储邻边的O。

public class Solution {
private static Queue<Integer> queue = null;
private static char[][] board;
private static int rows = 0;
private static int cols = 0;

public void solve(char[][] board) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if (board.length == 0 || board[0].length == 0) return;
queue = new LinkedList<Integer>();
board = board;
rows = board.length;
cols = board[0].length;
for (int i = 0; i < rows; i++) { // **important**
enqueue(i, 0);
enqueue(i, cols - 1);
}
for (int j = 1; j < cols - 1; j++) { // **important**
enqueue(0, j);
enqueue(rows - 1, j);
}
while (!queue.isEmpty()) {
int cur = queue.poll();
int x = cur / cols,
y = cur % cols;
if (board[x][y] == 'O') {
board[x][y] = 'D';
}
enqueue(x - 1, y);
enqueue(x + 1, y);
enqueue(x, y - 1);
enqueue(x, y + 1);
}
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (board[i][j] == 'D') board[i][j] = 'O';
else if (board[i][j] == 'O') board[i][j] = 'X';
}
}
queue = null;
board = null;
rows = 0;
cols = 0;
}
public static void enqueue(int x, int y) {
if (x >= 0 && x < rows && y >= 0 && y < cols && board[x][y] == 'O'){
queue.offer(x * cols + y);
}
}
}