# Number of Distinct Islands

`DFS`

Medium

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.

## Analysis & Solution

Shape一样，那么DFS搜索时的路径应该是一样的，那么反过来也成立。但是，有一个要注意的点是，DFS backtrack返回时，是因为碰到地图边缘了，还是因为旁边是海。这样就需要区别处理。

``````A = [[1,0],
[1,1]]
B = [[1,1],
[1,0]]``````

### DFS + HashSet (Hash by Path Signature)

``````class Solution {
public int numDistinctIslands(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
HashSet<String> islands = new HashSet<String>();
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) {
StringBuilder sb = new StringBuilder();
dfs(grid, sb, "o", i, j);
}
}
}

return islands.size();
}

private void dfs (int[][] grid, StringBuilder sb, String dir, int i, int j) {
if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0) {
return;
}

// mark the cell as visited
grid[i][j] = 0;
sb.append(dir);

dfs(grid, sb, "u", i - 1, j);
dfs(grid, sb, "l", i, j - 1);
dfs(grid, sb, "d", i + 1, j);
dfs(grid, sb, "r", i, j + 1);

sb.append("x");
}
}``````

### DFS + HashSet (Hash by Path Signature) Another One

``````public int numDistinctIslands(int[][] grid) {
Set<String> set = new HashSet<>();
for(int i = 0; i < grid.length; i++) {
for(int j = 0; j < grid[i].length; j++) {
if(grid[i][j] != 0) {
StringBuilder sb = new StringBuilder();
dfs(grid, i, j, sb, "o"); // origin
grid[i][j] = 0;
}
}
}
return set.size();
}
private void dfs(int[][] grid, int i, int j, StringBuilder sb, String dir) {
if(i < 0 || i == grid.length || j < 0 || j == grid[i].length
|| grid[i][j] == 0) return;
sb.append(dir);
grid[i][j] = 0;
dfs(grid, i-1, j, sb, "u");
dfs(grid, i+1, j, sb, "d");
dfs(grid, i, j-1, sb, "l");
dfs(grid, i, j+1, sb, "r");
sb.append("b"); // back
}``````

### DFS + Hash By Local Coordinates

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

public void explore(int r, int c, int r0, int c0) {
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 - r0) * 2 * grid[0].length + (c - c0));
explore(r+1, c, r0, c0);
explore(r-1, c, r0, c0);
explore(r, c+1, r0, c0);
explore(r, c-1, r0, c0);
}
}
public int numDistinctIslands(int[][] grid) {
this.grid = grid;
seen = new boolean[grid.length][grid[0].length];
Set shapes = new HashSet<HashSet<Integer>>();

for (int r = 0; r < grid.length; r++) {
for (int c = 0; c < grid[0].length; c++) {
shape = new HashSet<Integer>();
explore(r, c, r, c);
if (!shape.isEmpty()) {