[LeetCode/Java] Surrounded Regions

해석 적기

import java.util.LinkedList;
import java.util.Queue;

class Solution {
    static int[] dx = {1, 0, -1, 0};
    static int[] dy = {0, 1, 0, -1};

    public void solve(char[][] board) {
        int rows = board.length, cols = board[0].length;
        Queue<int[]> queue = new LinkedList<>();

        // Step 1: 경계에 있는 'O'를 찾아 큐에 추가하고 'T'로 변경
        for (int i = 0; i < rows; i++) {
            if (board[i][0] == 'O') bfs(board, i, 0, queue);
            if (board[i][cols - 1] == 'O') bfs(board, i, cols - 1, queue);
        }
        for (int j = 0; j < cols; j++) {
            if (board[0][j] == 'O') bfs(board, 0, j, queue);
            if (board[rows - 1][j] == 'O') bfs(board, rows - 1, j, queue);
        }

        // Step 2: 보드를 순회하며 'O'를 'X'로 변경, 'T'를 'O'로 복구
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == 'T') board[i][j] = 'O';
            }
        }
    }

    private void bfs(char[][] board, int x, int y, Queue<int[]> queue) {
        queue.add(new int[]{x, y});
        board[x][y] = 'T';

        while (!queue.isEmpty()) {
            int[] cur = queue.poll();
            for (int k = 0; k < 4; k++) {
                int nx = cur[0] + dx[k];
                int ny = cur[1] + dy[k];

                if (nx >= 0 && ny >= 0 && nx < board.length && ny < board[0].length && board[nx][ny] == 'O') {
                    board[nx][ny] = 'T'; // 경계와 연결된 'O'는 'T'로 변경
                    queue.add(new int[]{nx, ny});
                }
            }
        }
    }
}