### 题目

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. Note: The length of each dimension in the given grid does not exceed 50.

### union-find解连通图问题

11000
11000
00011
00011


4个1的偏移值分别为[0,0],[1,0],[0,1],[1,1]。所以偏移值序列为[0,0,1,0,0,1,1,1]。右下角岛的偏移值和左上角的一致，所以形状相同。

#### 代码

class Solution {

private static class UnionFind {

private int height;
private int width;
private int size;
private int[] board;

public UnionFind(int height, int width) {
this.height = height;
this.width = width;
size = height * width;
board = new int[size];
for (int i = 0; i < size; i++) board[i] = i;
}

public void kill(int a) {
board[a] = -1;
}

public void union(int a, int b) {
int rootA = find(a);
int rootB = find(b);
board[rootB] = rootA;
}

public int find(int n) {
if (board[n] == n) return n;
int root = find(board[n]);
board[n] = root; // path compression
return root;
}

public int indexOf(int x, int y) {
return x * width + y;
}

public int[] indexToPos(int idx) {
int[] pos = new int[2];
pos[0] = idx / width;
pos[1] = idx % width;
return pos;
}

public int count() {
Map<Integer, List<Integer>> shapeMap = new HashMap<>();
for (int i = 0; i < size; i++) {
if (board[i] >= 0) {
int root = find(i);
int[] pos = indexToPos(i);
int[] rootPos = indexToPos(root);
int[] offset = new int[]{pos[0] - rootPos[0], pos[1] - rootPos[1]};
List<Integer> shapeId = shapeMap.get(root);
}
}
Set<List<Integer>> shapeSet = new HashSet<>(shapeMap.values());
return shapeSet.size();
}

}

public int numDistinctIslands(int[][] grid) {
if (grid.length == 0) return 0;
int height = grid.length;
int width = grid[0].length;
UnionFind board = new UnionFind(height, width);
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
int idx = board.indexOf(i, j);
if (grid[i][j] == 1) {
if (i > 0 && grid[i - 1][j] == 1) board.union(idx, board.indexOf(i - 1, j));
if (j > 0 && grid[i][j - 1] ==1) board.union(idx, board.indexOf(i, j - 1));
} else {
board.kill(idx);
}
}
}
return board.count();
}
}


### DFS

#### 代码

class Solution {

public int numDistinctIslands(int[][] grid) {
if (grid.length == 0) return 0;
height = grid.length;
width = grid[0].length;
localGrid = grid;
Set<String> islands = new HashSet<>();
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
if (localGrid[i][j] == 1) {
}
}
}
return islands.size();
}

private static final int[][] dirs = new int[][]{
{0, 1},
{1, 0},
{0, -1},
{-1, 0}
};
private int height;
private int width;
private int[][] localGrid;

private StringBuilder dfs(int x, int y, int offsetX, int offsetY) {
StringBuilder sb = new StringBuilder();
if (x >= 0 && x < height && y >= 0 && y < width && localGrid[x][y] == 1) {
sb.append(offsetX).append(offsetY);
localGrid[x][y] = 0;
for (int[] dir : dirs) sb.append(dfs(x + dir[0], y + dir[1], offsetX + dir[0], offsetY + dir[1]));
}
return sb;
}

}