694. Number of Distinct Islands

Problem:

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

Given the above grid map, return 3.

Notice that:

11
1

and

 1
11

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

Analysis:
The tricky part is how to make same shape the same. We store the diff of each point in the shape by x-axial and y axial separately from the start to a list. 

11000
11000
00011
00011

The coordinates of last three points of the first shape are: (0, 1), (1,0), (1, 1) the base is (0, 0)
The diff are: 0,1,1,0,1,1
The coordinates of last three points of the second shape are: (2,4), (3,3), (3,4) the base (2,3)
The diff are: 0,1,1,0,1,1
We can draw the conclusion that if the diff lists are the same, then the shape are the same. 

Solution:

class Solution {
    public int numDistinctIslands(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0)
            return 0;
        Set<List<Integer>> set = new HashSet<>();
        int[][] dirs = new int[][]{{1, 0}, {-1 ,0}, {0, 1}, {0, -1}};
        int m = grid.length;
        int n = grid[0].length;
        for (int  i = 0; i < m; i++) 
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    Queue<Integer> queue = new LinkedList<>();
                    queue.offer(i*n + j);
                    grid[i][j] = -1;
                    List<Integer> list = new ArrayList<>();
                    while (!queue.isEmpty()) {
                        int cur = queue.poll();
                        for (int[] dir: dirs) {
                            int newr = cur/n + dir[0];
                            int newc = cur%n + dir[1];
                            if (newr >= 0 && newr < m && newc >=0 && newc < n && grid[newr][newc] == 1) {
                                grid[newr][newc] = -1;
                                queue.offer(newr*n + newc);
                                list.add(newr - i);
                                list.add(newc - j);
                            }
                        }
                    }
                    set.add(list);
                }
            }
        return set.size();
    }
}