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给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 '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。
- 输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
- 输出：[["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。
这道题目和1020. 飞地的数量正好反过来了，1020. 飞地的数量是求 地图中间的空格数，而本题是要把地图中间的'O'都改成'X'。
那么两题在思路上也是差不多的。
依然是从地图周边出发，将周边空格相邻的'O'都做上标记，然后在遍历一遍地图，遇到 'O' 且没做过标记的，那么都是地图中间的'O'，全部改成'X'就行。
有的录友可能想，我在定义一个 visited 二维数组，单独标记周边的'O'，然后遍历地图的时候同时对 数组board 和 数组visited 进行判断，是否'O'改成'X'。
这样做其实就有点麻烦了，不用额外定义空间了，标记周边的'O'，可以直接改board的数值为其他特殊值。
步骤一：深搜或者广搜将地图周边的'O'全部改成'A'，如图所示：
步骤二：在遍历地图，将'O'全部改成'X'（地图中间的'O'改成了'X'），将'A'改回'O'（保留的地图周边的'O'），如图所示：
整体C++代码如下，以下使用dfs实现，其实遍历方式dfs，bfs都是可以的。
class Solution {
private:
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
void dfs(vector<vector<char>>& board, int x, int y) {
board[x][y] = 'A';
for (int i = 0; i < 4; i++) { // 向四个方向遍历
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
// 超过边界
if (nextx < 0 || nextx >= board.size() || nexty < 0 || nexty >= board[0].size()) continue;
// 不符合条件，不继续遍历
if (board[nextx][nexty] == 'X' || board[nextx][nexty] == 'A') continue;
dfs (board, nextx, nexty);
}
return;
}
public:
void solve(vector<vector<char>>& board) {
int n = board.size(), m = board[0].size();
// 步骤一：
// 从左侧边，和右侧边 向中间遍历
for (int i = 0; i < n; i++) {
if (board[i][0] == 'O') dfs(board, i, 0);
if (board[i][m - 1] == 'O') dfs(board, i, m - 1);
}
// 从上边和下边 向中间遍历
for (int j = 0; j < m; j++) {
if (board[0][j] == 'O') dfs(board, 0, j);
if (board[n - 1][j] == 'O') dfs(board, n - 1, j);
}
// 步骤二：
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == 'O') board[i][j] = 'X';
if (board[i][j] == 'A') board[i][j] = 'O';
}
}
}
};// 广度优先遍历
// 使用 visited 数组进行标记
class Solution {
private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 四个方向
public void solve(char[][] board) {
// rowSize：行的长度，colSize：列的长度
int rowSize = board.length, colSize = board[0].length;
boolean[][] visited = new boolean[rowSize][colSize];
Queue<int[]> queue = new ArrayDeque<>();
// 从左侧边，和右侧边遍历
for (int row = 0; row < rowSize; row++) {
if (board[row][0] == 'O') {
visited[row][0] = true;
queue.add(new int[]{row, 0});
}
if (board[row][colSize - 1] == 'O') {
visited[row][colSize - 1] = true;
queue.add(new int[]{row, colSize - 1});
}
}
// 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
// 所以在遍历上边和下边时可以不用遍历四个角
for (int col = 1; col < colSize - 1; col++) {
if (board[0][col] == 'O') {
visited[0][col] = true;
queue.add(new int[]{0, col});
}
if (board[rowSize - 1][col] == 'O') {
visited[rowSize - 1][col] = true;
queue.add(new int[]{rowSize - 1, col});
}
}
// 广度优先遍历，把没有被 'X' 包围的 'O' 进行标记
while (!queue.isEmpty()) {
int[] current = queue.poll();
for (int[] pos: position) {
int row = current[0] + pos[0], col = current[1] + pos[1];
// 如果范围越界、位置已被访问过、该位置的值不是 'O'，就直接跳过
if (row < 0 || row >= rowSize || col < 0 || col >= colSize) continue;
if (visited[row][col] || board[row][col] != 'O') continue;
visited[row][col] = true;
queue.add(new int[]{row, col});
}
}
// 遍历数组，把没有被标记的 'O' 修改成 'X'
for (int row = 0; row < rowSize; row++) {
for (int col = 0; col < colSize; col++) {
if (board[row][col] == 'O' && !visited[row][col]) board[row][col] = 'X';
}
}
}
}// 广度优先遍历
// 直接修改 board 的值为其他特殊值
class Solution {
private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 四个方向
public void solve(char[][] board) {
// rowSize：行的长度，colSize：列的长度
int rowSize = board.length, colSize = board[0].length;
Queue<int[]> queue = new ArrayDeque<>();
// 从左侧边，和右侧边遍历
for (int row = 0; row < rowSize; row++) {
if (board[row][0] == 'O')
queue.add(new int[]{row, 0});
if (board[row][colSize - 1] == 'O')
queue.add(new int[]{row, colSize - 1});
}
// 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
// 所以在遍历上边和下边时可以不用遍历四个角
for (int col = 1; col < colSize - 1; col++) {
if (board[0][col] == 'O')
queue.add(new int[]{0, col});
if (board[rowSize - 1][col] == 'O')
queue.add(new int[]{rowSize - 1, col});
}
// 广度优先遍历，把没有被 'X' 包围的 'O' 修改成特殊值
while (!queue.isEmpty()) {
int[] current = queue.poll();
board[current[0]][current[1]] = 'A';
for (int[] pos: position) {
int row = current[0] + pos[0], col = current[1] + pos[1];
// 如果范围越界、该位置的值不是 'O'，就直接跳过
if (row < 0 || row >= rowSize || col < 0 || col >= colSize) continue;
if (board[row][col] != 'O') continue;
queue.add(new int[]{row, col});
}
}
// 遍历数组，把 'O' 修改成 'X'，特殊值修改成 'O'
for (int row = 0; row < rowSize; row++) {
for (int col = 0; col < colSize; col++) {
if (board[row][col] == 'A') board[row][col] = 'O';
else if (board[row][col] == 'O') board[row][col] = 'X';
}
}
}
}//BFS（使用helper function）
class Solution {
int[][] dir ={{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public void solve(char[][] board) {
for(int i = 0; i < board.length; i++){
if(board[i][0] == 'O') bfs(board, i, 0);
if(board[i][board[0].length - 1] == 'O') bfs(board, i, board[0].length - 1);
}
for(int j = 1 ; j < board[0].length - 1; j++){
if(board[0][j] == 'O') bfs(board, 0, j);
if(board[board.length - 1][j] == 'O') bfs(board, board.length - 1, j);
}
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == 'O') board[i][j] = 'X';
if(board[i][j] == 'A') board[i][j] = 'O';
}
}
}
private void bfs(char[][] board, int x, int y){
Queue<Integer> que = new LinkedList<>();
board[x][y] = 'A';
que.offer(x);
que.offer(y);
while(!que.isEmpty()){
int currX = que.poll();
int currY = que.poll();
for(int i = 0; i < 4; i++){
int nextX = currX + dir[i][0];
int nextY = currY + dir[i][1];
if(nextX < 0 || nextY < 0 || nextX >= board.length || nextY >= board[0].length)
continue;
if(board[nextX][nextY] == 'X'|| board[nextX][nextY] == 'A')
continue;
bfs(board, nextX, nextY);
}
}
}
}// 深度优先遍历
// 使用 visited 数组进行标记
class Solution {
private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 四个方向
public void dfs(char[][] board, int row, int col, boolean[][] visited) {
for (int[] pos: position) {
int nextRow = row + pos[0], nextCol = col + pos[1];
// 位置越界
if (nextRow < 0 || nextRow >= board.length || nextCol < 0 || nextCol >= board[0].length)
continue;
// 位置已被访问过、新位置值不是 'O'
if (visited[nextRow][nextCol] || board[nextRow][nextCol] != 'O') continue;
visited[nextRow][nextCol] = true;
dfs(board, nextRow, nextCol, visited);
}
}
public void solve(char[][] board) {
int rowSize = board.length, colSize = board[0].length;
boolean[][] visited = new boolean[rowSize][colSize];
// 从左侧遍、右侧遍遍历
for (int row = 0; row < rowSize; row++) {
if (board[row][0] == 'O' && !visited[row][0]) {
visited[row][0] = true;
dfs(board, row, 0, visited);
}
if (board[row][colSize - 1] == 'O' && !visited[row][colSize - 1]) {
visited[row][colSize - 1] = true;
dfs(board, row, colSize - 1, visited);
}
}
// 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
// 所以在遍历上边和下边时可以不用遍历四个角
for (int col = 1; col < colSize - 1; col++) {
if (board[0][col] == 'O' && !visited[0][col]) {
visited[0][col] = true;
dfs(board, 0, col, visited);
}
if (board[rowSize - 1][col] == 'O' && !visited[rowSize - 1][col]) {
visited[rowSize - 1][col] = true;
dfs(board, rowSize - 1, col, visited);
}
}
// 遍历数组，把没有被标记的 'O' 修改成 'X'
for (int row = 0; row < rowSize; row++) {
for (int col = 0; col < colSize; col++) {
if (board[row][col] == 'O' && !visited[row][col]) board[row][col] = 'X';
}
}
}
}// 深度优先遍历
// // 直接修改 board 的值为其他特殊值
class Solution {
private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 四个方向
public void dfs(char[][] board, int row, int col) {
for (int[] pos: position) {
int nextRow = row + pos[0], nextCol = col + pos[1];
// 位置越界
if (nextRow < 0 || nextRow >= board.length || nextCol < 0 || nextCol >= board[0].length)
continue;
// 新位置值不是 'O'
if (board[nextRow][nextCol] != 'O') continue;
board[nextRow][nextCol] = 'A'; // 修改为特殊值
dfs(board, nextRow, nextCol);
}
}
public void solve(char[][] board) {
int rowSize = board.length, colSize = board[0].length;
// 从左侧遍、右侧遍遍历
for (int row = 0; row < rowSize; row++) {
if (board[row][0] == 'O') {
board[row][0] = 'A';
dfs(board, row, 0);
}
if (board[row][colSize - 1] == 'O') {
board[row][colSize - 1] = 'A';
dfs(board, row, colSize - 1);
}
}
// 从上边和下边遍历，在对左侧边和右侧边遍历时我们已经遍历了矩阵的四个角
// 所以在遍历上边和下边时可以不用遍历四个角
for (int col = 1; col < colSize - 1; col++) {
if (board[0][col] == 'O') {
board[0][col] = 'A';
dfs(board, 0, col);
}
if (board[rowSize - 1][col] == 'O') {
board[rowSize - 1][col] = 'A';
dfs(board, rowSize - 1, col);
}
}
// 遍历数组，把 'O' 修改成 'X'，特殊值修改成 'O'
for (int row = 0; row < rowSize; row++) {
for (int col = 0; col < colSize; col++) {
if (board[row][col] == 'O') board[row][col] = 'X';
else if (board[row][col] == 'A') board[row][col] = 'O';
}
}
}
}//DFS(有終止條件)
class Solution {
int[][] dir ={{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public void solve(char[][] board) {
for(int i = 0; i < board.length; i++){
if(board[i][0] == 'O') dfs(board, i, 0);
if(board[i][board[0].length - 1] == 'O') dfs(board, i, board[0].length - 1);
}
for(int j = 1 ; j < board[0].length - 1; j++){
if(board[0][j] == 'O') dfs(board, 0, j);
if(board[board.length - 1][j] == 'O') dfs(board, board.length - 1, j);
}
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == 'O') board[i][j] = 'X';
if(board[i][j] == 'A') board[i][j] = 'O';
}
}
}
private void dfs(char[][] board, int x, int y){
if(board[x][y] == 'X'|| board[x][y] == 'A')
return;
board[x][y] = 'A';
for(int i = 0; i < 4; i++){
int nextX = x + dir[i][0];
int nextY = y + dir[i][1];
if(nextX < 0 || nextY < 0 || nextX >= board.length || nextY >= board[0].length)
continue;
// if(board[nextX][nextY] == 'X'|| board[nextX][nextY] == 'A')
// continue;
dfs(board, nextX, nextY);
}
}
}// 深度优先遍历
class Solution:
dir_list = [(0, 1), (0, -1), (1, 0), (-1, 0)]
def solve(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
row_size = len(board)
column_size = len(board[0])
visited = [[False] * column_size for _ in range(row_size)]
# 从边缘开始，将边缘相连的O改成A。然后遍历所有，将A改成O，O改成X
# 第一行和最后一行
for i in range(column_size):
if board[0][i] == "O" and not visited[0][i]:
self.dfs(board, 0, i, visited)
if board[row_size-1][i] == "O" and not visited[row_size-1][i]:
self.dfs(board, row_size-1, i, visited)
# 第一列和最后一列
for i in range(1, row_size-1):
if board[i][0] == "O" and not visited[i][0]:
self.dfs(board, i, 0, visited)
if board[i][column_size-1] == "O" and not visited[i][column_size-1]:
self.dfs(board, i, column_size-1, visited)
for i in range(row_size):
for j in range(column_size):
if board[i][j] == "A":
board[i][j] = "O"
elif board[i][j] == "O":
board[i][j] = "X"
def dfs(self, board, x, y, visited):
if visited[x][y] or board[x][y] == "X":
return
visited[x][y] = True
board[x][y] = "A"
for i in range(4):
new_x = x + self.dir_list[i][0]
new_y = y + self.dir_list[i][1]
if new_x >= len(board) or new_y >= len(board[0]) or new_x < 0 or new_y < 0:
continue
self.dfs(board, new_x, new_y, visited)/**
* @description 深度搜索优先
* @param {character[][]} board
* @return {void} Do not return anything, modify board in-place instead.
*/
function solve(board) {
const dir = [[-1, 0], [1, 0], [0, -1], [0, 1]];
const [rowSize, colSize] = [board.length, board[0].length];
function dfs(board, x, y) {
board[x][y] = 'A';
for (let i = 0; i < 4; i++) {
const nextX = dir[i][0] + x;
const nextY = dir[i][1] + y;
if (nextX < 0 || nextX >= rowSize || nextY < 0 || nextY >= colSize) {
continue;
}
if (board[nextX][nextY] === 'O') {
dfs(board, nextX, nextY);
}
}
}
for (let i = 0; i < rowSize; i++) {
if (board[i][0] === 'O') {
dfs(board, i, 0);
}
if (board[i][colSize - 1] === 'O') {
dfs(board, i, colSize - 1);
}
}
for (let i = 1; i < colSize - 1; i++) {
if (board[0][i] === 'O') {
dfs(board, 0, i);
}
if (board[rowSize - 1][i] === 'O') {
dfs(board, rowSize - 1, i);
}
}
for (let i = 0; i < rowSize; i++) {
for (let k = 0; k < colSize; k++) {
if (board[i][k] === 'A') {
board[i][k] = 'O';
} else if (board[i][k] === 'O') {
board[i][k] = 'X';
}
}
}
}
/**
* @description 广度搜索优先
* @param {character[][]} board
* @return {void} Do not return anything, modify board in-place instead.
*/
function solve(board) {
const dir = [[-1, 0], [1, 0], [0, -1], [0, 1]];
const [rowSize, colSize] = [board.length, board[0].length];
function bfs(board, x, y) {
board[x][y] = 'A';
const stack = [y, x];
while (stack.length !== 0) {
const top = [stack.pop(), stack.pop()];
for (let i = 0; i < 4; i++) {
const nextX = dir[i][0] + top[0];
const nextY = dir[i][1] + top[1];
if (nextX < 0 || nextX >= rowSize || nextY < 0 || nextY >= colSize) {
continue;
}
if (board[nextX][nextY] === 'O') {
board[nextX][nextY] = 'A';
stack.push(nextY, nextX);
}
}
}
for (let i = 0; i < 4; i++) {
const nextX = dir[i][0] + x;
const nextY = dir[i][1] + y;
if (nextX < 0 || nextX >= rowSize || nextY < 0 || nextY >= colSize) {
continue;
}
if (board[nextX][nextY] === 'O') {
dfs(board, nextX, nextY);
}
}
}
for (let i = 0; i < rowSize; i++) {
if (board[i][0] === 'O') {
bfs(board, i, 0);
}
if (board[i][colSize - 1] === 'O') {
bfs(board, i, colSize - 1);
}
}
for (let i = 1; i < colSize - 1; i++) {
if (board[0][i] === 'O') {
bfs(board, 0, i);
}
if (board[rowSize - 1][i] === 'O') {
bfs(board, rowSize - 1, i);
}
}
for (let i = 0; i < rowSize; i++) {
for (let k = 0; k < colSize; k++) {
if (board[i][k] === 'A') {
board[i][k] = 'O';
} else if (board[i][k] === 'O') {
board[i][k] = 'X';
}
}
}
}dfs:
var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
func solve(board [][]byte) {
rows, cols := len(board), len(board[0])
// 列
for i := 0; i < rows; i++ {
if board[i][0] == 'O' {
dfs(board, i, 0)
}
if board[i][cols-1] == 'O' {
dfs(board, i, cols-1)
}
}
// 行
for j := 0; j < cols; j++ {
if board[0][j] == 'O' {
dfs(board, 0, j)
}
if board[rows-1][j] == 'O' {
dfs(board, rows-1, j)
}
}
for _, r := range board {
for j, c := range r {
if c == 'A' {
r[j] = 'O'
continue
}
if c == 'O' {
r[j] = 'X'
}
}
}
}
func dfs(board [][]byte, i, j int) {
board[i][j] = 'A'
for _, d := range DIRECTIONS {
x, y := i+d[0], j+d[1]
if x < 0 || x >= len(board) || y < 0 || y >= len(board[0]) {
continue
}
if board[x][y] == 'O' {
dfs(board, x, y)
}
}
}bfs:
var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
func solve(board [][]byte) {
rows, cols := len(board), len(board[0])
// 列
for i := 0; i < rows; i++ {
if board[i][0] == 'O' {
bfs(board, i, 0)
}
if board[i][cols-1] == 'O' {
bfs(board, i, cols-1)
}
}
// 行
for j := 0; j < cols; j++ {
if board[0][j] == 'O' {
bfs(board, 0, j)
}
if board[rows-1][j] == 'O' {
bfs(board, rows-1, j)
}
}
for _, r := range board {
for j, c := range r {
if c == 'A' {
r[j] = 'O'
continue
}
if c == 'O' {
r[j] = 'X'
}
}
}
}
func bfs(board [][]byte, i, j int) {
queue := [][]int{{i, j}}
board[i][j] = 'A'
for len(queue) > 0 {
cur := queue[0]
queue = queue[1:]
for _, d := range DIRECTIONS {
x, y := cur[0]+d[0], cur[1]+d[1]
if x < 0 || x >= len(board) || y < 0 || y >= len(board[0]) {
continue
}
if board[x][y] == 'O' {
board[x][y] = 'A'
queue = append(queue, []int{x, y})
}
}
}
}bfs:
impl Solution {
const DIRECTIONS: [(isize, isize); 4] = [(0, 1), (0, -1), (1, 0), (-1, 0)];
pub fn solve(board: &mut Vec<Vec<char>>) {
let (rows, cols) = (board.len(), board[0].len());
// 列
for i in 0..rows {
if board[i][0] == 'O' {
Self::dfs(board, i, 0);
}
if board[i][cols - 1] == 'O' {
Self::dfs(board, i, cols - 1);
}
}
//行
for j in 0..cols {
if board[0][j] == 'O' {
Self::dfs(board, 0, j);
}
if board[rows - 1][j] == 'O' {
Self::dfs(board, rows - 1, j);
}
}
for v in board.iter_mut() {
for c in v.iter_mut() {
if *c == 'A' {
*c = 'O';
continue;
}
if *c == 'O' {
*c = 'X';
}
}
}
}
pub fn dfs(board: &mut [Vec<char>], i: usize, j: usize) {
board[i][j] = 'A';
for (d1, d2) in Self::DIRECTIONS {
let (x, y) = (i as isize + d1, j as isize + d2);
if x < 0 || x >= board.len() as isize || y < 0 || y >= board[0].len() as isize {
continue;
}
let (x, y) = (x as usize, y as usize);
if board[x][y] == 'O' {
Self::dfs(board, x, y);
}
}
}
}bfs:
use std::collections::VecDeque;
impl Solution {
const DIRECTIONS: [(isize, isize); 4] = [(0, 1), (0, -1), (1, 0), (-1, 0)];
pub fn solve(board: &mut Vec<Vec<char>>) {
let (rows, cols) = (board.len(), board[0].len());
// 列
for i in 0..rows {
if board[i][0] == 'O' {
Self::bfs(board, i, 0);
}
if board[i][cols - 1] == 'O' {
Self::bfs(board, i, cols - 1);
}
}
//行
for j in 0..cols {
if board[0][j] == 'O' {
Self::bfs(board, 0, j);
}
if board[rows - 1][j] == 'O' {
Self::bfs(board, rows - 1, j);
}
}
for v in board.iter_mut() {
for c in v.iter_mut() {
if *c == 'A' {
*c = 'O';
continue;
}
if *c == 'O' {
*c = 'X';
}
}
}
}
pub fn bfs(board: &mut [Vec<char>], i: usize, j: usize) {
let mut queue = VecDeque::from([(i, j)]);
board[i][j] = 'A';
while let Some((i, j)) = queue.pop_front() {
for (d1, d2) in Self::DIRECTIONS {
let (x, y) = (i as isize + d1, j as isize + d2);
if x < 0 || x >= board.len() as isize || y < 0 || y >= board[0].len() as isize {
continue;
}
let (x, y) = (x as usize, y as usize);
if board[x][y] == 'O' {
board[x][y] = 'A';
queue.push_back((x, y));
}
}
}
}
}
