Number of Distinct Islands

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

Example 1:

11000
11000
00011
00011

Given the above grid map, return 1.

Example 2:

11011
10000
00001
11011

Idea: use string to represent the moving state, then each shape is the same, do dfs, then the string should be the same. Note that a "b" should be added at the end

Because directly down-> left, it is not equivalent to down-> deadend-> back-> left; these are two different moving methods;

class Solution {
    public int numDistinctIslands(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        
        int n = grid.length;
        int m = grid[0].length;
        HashSet<String> set = new HashSet<String>();
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == 1) {
                    StringBuilder sb = new StringBuilder();
                    dfs(grid, sb, i, j, "o");
                    set.add(sb.toString());
                }
            }
        }
        return set.size();
    }
    
    private int[] dx = {0,0,-1,1};
    private int[] dy = {-1,1,0,0};
    
    private void dfs(int[][] grid, StringBuilder sb, int x, int y, String direction) {
        if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
            return;
        }
        if(grid[x][y] == 1) {
            grid[x][y] = 0;
            sb.append(direction);
            dfs(grid, sb, x + 1, y, "d");
            dfs(grid, sb, x - 1, y, "u");
            dfs(grid, sb, x, y + 1, "r");
            dfs(grid, sb, x, y - 1, "l");
            sb.append("b"); // 因为直接down -> left,不等同于down-> deadend -> back -> left;
        }
    }
}