不同岛屿的数量

/*
 * 题目思路：
 * 使用Java Stream API实现找到不同岛屿数量的方法稍显复杂，因为Stream更适合处理非递归的、无状态的操作。
 * 但我们依然可以借助Stream来进行初始化、遍历，并在其中调用DFS方法。
 */
import java.util.*;
import java.util.stream.*;
 
public class NumberOfDistinctIslandsStream {
    private static final int[] dx = {0, 1, 0, -1};
    private static final int[] dy = {1, 0, -1, 0};
 
    public int numDistinctIslands(int[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        Set<String> islands = new HashSet<>();
        boolean[][] visited = new boolean[grid.length][grid[0].length];
 
        IntStream.range(0, grid.length).forEach(i ->
            IntStream.range(0, grid[0].length).forEach(j -> {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    StringBuilder shape = new StringBuilder();
                    dfs(grid, visited, i, j, shape, 'o');
                    islands.add(shape.toString());
                }
            })
        );
        return islands.size();
    }
 
    private void dfs(int[][] grid, boolean[][] visited, int x, int y, StringBuilder shape, char dir) {
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == 0 || visited[x][y]) return;
        visited[x][y] = true;
        shape.append(dir);
        for (int i = 0; i < 4; i++) {
            dfs(grid, visited, x + dx[i], y + dy[i], shape, (char) ('0' + i));
        }
        shape.append('b');
    }
}