Number of Distinct Islands

Given a non-empty 2D arraygridof 0's and 1's, an island is a group of1'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

Example 1:

11000
11000
00011
00011

Given the above grid map, return1.

Example 2:

11011
10000
00001
11011

Given the above grid map, return3.

Notice that:

11
1

and

 1
11

are considered different island shapes, because we do not consider reflection / rotation.

Note

利用BFS暴力去做, 找到岛屿单元的相对坐标(the entry point)， 使用hashset， set 里面是BFS返回的一系列坐标

给横纵坐标分别一个队列

Code

class Solution {

    final int[] dx = {0, 0, 1, -1};
    final int[] dy = {1, -1, 0, 0};
    public int numDistinctIslands(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }

        Set<List<List<Integer>>> set = new HashSet<>();
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j <grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    List<List<Integer>> list = bfs(grid, i, j);
                    set.add(list);
                }
            }
        }
        return set.size();
    }

    private List<List<Integer>> bfs(int[][] grid, int a, int b) {
        List<List<Integer>> res = new ArrayList<>();
        int m = grid.length, n = grid[0].length;
        Queue<Integer> qx = new LinkedList<>();
        Queue<Integer> qy = new LinkedList<>();
        qx.offer(a);
        qy.offer(b);
        while (!qx.isEmpty() && !qy.isEmpty()) {
            int x = qx.poll();
            int y = qy.poll();
            List<Integer> list = new ArrayList<>();
            list.add(x - a);
            list.add(y - b);
            res.add(list);
            for (int i = 0; i < 4; i++) {
                int nx = x + dx[i];
                int ny = y + dy[i];
                if (nx >= 0 &&nx < m && ny >= 0 && ny < n && grid[nx][ny] == 1) {
                    grid[nx][ny] = 0;
                    qx.offer(nx);
                    qy.offer(ny);
                }
            }
        }
        return res;
    }
}