【题目】*130. 被围绕的区域
给定一个二维的矩阵，包含 ‘X’ 和 ‘O’（字母 O）。
找到所有被 ‘X’ 围绕的区域，并将这些区域里所有的 ‘O’ 用 ‘X’ 填充。
示例:
X X X X
X O O X
X X O X
X O X X
运行你的函数后，矩阵变为：
X X X X
X X X X
X X X X
X O X X
解释:
被围绕的区间不会存在于边界上，换句话说，任何边界上的 ‘O’ 都不会被填充为 ‘X’。 任何不在边界上，或不与边界上的 ‘O’ 相连的 ‘O’ 最终都会被填充为 ‘X’。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
【解题思路1】DFS
从边缘入手，把边缘以及和边缘相连的O标记，比如改为#或A等
最后再遍历一遍，把标记的格子改回O，把O改为X
class Solution {
int n, m;
public void solve(char[][] board) {
n = board.length;
if (n == 0) {
return;
}
m = board[0].length;
for (int i = 0; i < n; i++) {
dfs(board, i, 0);
dfs(board, i, m - 1);
}
for (int i = 1; i < m - 1; i++) {
dfs(board, 0, i);
dfs(board, n - 1, i);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == 'A') {
board[i][j] = 'O';
} else if (board[i][j] == 'O') {
board[i][j] = 'X';
}
}
}
}
public void dfs(char[][] board, int x, int y) {
if (x < 0 || x >= n || y < 0 || y >= m || board[x][y] != 'O') {
return;
}
board[x][y] = 'A';
dfs(board, x + 1, y);
dfs(board, x - 1, y);
dfs(board, x, y + 1);
dfs(board, x, y - 1);
}
}
【解题思路2】BFS
class Solution {
int[] dx = {
1, -1, 0, 0};
int[] dy = {
0, 0, 1, -1};
public void solve(char[][] board) {
int n = board.length;
if (n == 0) {
return;
}
int m = board[0].length;
Queue<int[]> queue = new LinkedList<int[]>();
for (int i = 0; i < n; i++) {
if (board[i][0] == 'O') {
queue.offer(new int[]{
i, 0});
}
if (board[i][m - 1] == 'O') {
queue.offer(new int[]{
i, m - 1});
}
}
for (int i = 1; i < m - 1; i++) {
if (board[0][i] == 'O') {
queue.offer(new int[]{
0, i});
}
if (board[n - 1][i] == 'O') {
queue.offer(new int[]{
n - 1, i});
}
}
while (!queue.isEmpty()) {
int[] cell = queue.poll();
int x = cell[0], y = cell[1];
board[x][y] = 'A';
for (int i = 0; i < 4; i++) {
int mx = x + dx[i], my = y + dy[i];
if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O') {
continue;
}
queue.offer(new int[]{
mx, my});
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == 'A') {
board[i][j] = 'O';
} else if (board[i][j] == 'O') {
board[i][j] = 'X';
}
}
}
}
}