711 Number of Distinct Islands II

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 they have the same shape, or have the same shape after rotation (90, 180, or 270 degrees only) or reflection (left/right direction or up/down direction).

Example 1:

11000
10000
00001
00011

Given the above grid map, return 1. Notice that:

11
1

and

 1
11

are considered same island shapes. Because if we make a 180 degrees clockwise rotation on the first island, then two islands will have the same shapes.

Example 2:

11100
10001
01001
01110

Given the above grid map, return 2. Here are the two distinct islands:

111
1

and

1
1

Notice that:

111
1

and

1
111

are considered same island shapes. Because if we flip the first array in the up/down direction, then they have the same shapes.

Note: The length of each dimension in the given grid does not exceed 50.Accepted6,691Submissions13,589

这题的其实跟694 Number of Distinct Islands思路一样。那么难题是怎么吧reflection和rotation的island编码呢？自己是想不出来了，看答案也看了半天。其实，rotation就是把现在的（x, y)做一下变换：x y, x -y, -x y, -x -y；reflection就是把现在的左边做以下变换：y x, y -x, -y x, -y -x。我们把组成island的每一个点都做一遍这样的变换，找出8种形态。每个形态，我们要loop一遍，算一下这个岛跟最左上一点的差值。然后排序，用所产生的数字最小值作为key。至于为什么这个编码过程是work的，就不太懂了。T:O(r * clog(r * c)),这个log是因为要给8种形态排序。S:O(r * c)，递归深度

class Solution {
	int[][] grid;
	boolean[][] seen;
	ArrayList<Integer> shape;

	public void explore(int r, int c) {
		if (0 <= r && r < grid.length && 0 <= c && c < grid[0].length && grid[r][c] == 1 && !seen[r][c]) {
			seen[r][c] = true;
			shape.add(r * grid[0].length + c);
			explore(r + 1, c);
			explore(r - 1, c);
			explore(r, c + 1);
			explore(r, c - 1);
		}
	}
	// x y, x -y, -x y, -x -y, --- rotation
	// y x, y -x, -y x, -y -x --- reflection 
	int[] xMul = {1, 1, -1, -1, 1, -1, 1, -1};
	int[] yMul = {1, -1, 1, -1, 1, 1, -1, -1};
	public String canonical(ArrayList<Integer> shape) {
		String ans = "";
		int lift = grid.length + grid[0].length;
		int[] out = new int[shape.size()];
		int[] xs = new int[shape.size()];
		int[] ys = new int[shape.size()];

		for (int c = 0; c < 8; ++c) {
			int t = 0;
			for (int z : shape) {
				int x = z / grid[0].length;
				int y = z % grid[0].length;

				if (c <= 3) {
					xs[t] = x * xMul[c];
					ys[t] = y * yMul[c];
				} else {
					xs[t] = y * xMul[c];
					ys[t] = x * yMul[c];
				}
				t++;
			}

			int mx = xs[0], my = ys[0];
			for (int x : xs)
				mx = Math.min(mx, x);
			for (int y : ys)
				my = Math.min(my, y);

			for (int j = 0; j < shape.size(); ++j) {
				out[j] = (xs[j] - mx) * lift + (ys[j] - my);
			}
			Arrays.sort(out);
			String candidate = Arrays.toString(out);
			if (ans.compareTo(candidate) < 0)
				ans = candidate;
		}
		return ans;
	}

	public int numDistinctIslands2(int[][] grid) {
		this.grid = grid;
		seen = new boolean[grid.length][grid[0].length];
		Set shapes = new HashSet<String>();

		for (int r = 0; r < grid.length; ++r) {
			for (int c = 0; c < grid[0].length; ++c) {
				shape = new ArrayList();
				explore(r, c);
				if (!shape.isEmpty()) {
					shapes.add(canonical(shape));
				}
			}
		}

		return shapes.size();
	}
}