边界着色

/*
题目思路：
1. 获取初始位置的颜色，检查如果初始颜色和目标颜色相同，直接返回。
2. 使用广度优先搜索（BFS）找到与初始位置颜色相同的连通分量。
3. 使用Java Stream API来遍历和操作数组数据。
*/
import java.util.*;
import java.util.stream.*;

public class Solution {
    private int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    public int[][] colorBorder(int[][] grid, int row, int col, int color) {
        int originalColor = grid[row][col];
        if (originalColor == color) return grid;

        boolean[][] visited = new boolean[grid.length][grid[0].length];
        Queue<int[]> queue = new LinkedList<>();
        queue.add(new int[]{row, col});
        visited[row][col] = true;
        List<int[]> borderCells = new ArrayList<>();

        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int r = cell[0], c = cell[1];
            boolean isBorder = false;
            for (int[] dir : directions) {
                int newRow = r + dir[0];
                int newCol = c + dir[1];
                if (newRow < 0 || newRow >= grid.length || newCol < 0 || newCol >= grid[0].length || grid[newRow][newCol] != originalColor) {
                    isBorder = true;
                } else if (!visited[newRow][newCol]) {
                    queue.add(new int[]{newRow, newCol});
                    visited[newRow][newCol] = true;
                }
            }
            if (isBorder) borderCells.add(new int[]{r, c});
        }

        borderCells.stream().forEach(cell -> grid[cell[0]][cell[1]] = color);

        return grid;
    }
}