# 711 Number of Distinct Islands II

## 1. Question

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.

## 2. Implementation

(1) DFS

1.首先通过DFS寻找island，并在这个过程中通过一个ArrayList存DFS找到的一个island的所有坐标
2.通过normalize(),得到当前island的shape key。normalize会将island的每个坐标转换成8种变化，然后通过排序得到取第一个作为表示该island形状的key，将key放入set中
3.最后set的size就代表有多少个distinct island
class Solution {
public int numDistinctIslands2(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
Set<String> set = new HashSet();
int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int[][] transformations = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
int m = grid.length, n = grid[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
List<int[]> cells = new ArrayList();
getIslandsByDFS(i, j, grid, cells, directions);
String key = normalize(cells, transformations);
}
}
}
return set.size();
}
public void getIslandsByDFS(int row, int col, int[][] grid, List<int[]> cells, int[][] directions) {
grid[row][col] = 0;
for (int[] direction : directions) {
int nextRow = row + direction[0];
int nextCol = col + direction[1];
if (isValid(nextRow, nextCol, grid)) {
getIslandsByDFS(nextRow, nextCol, grid, cells, directions);
}
}
}
public boolean isValid(int row, int col, int[][] grid) {
return row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && grid[row][col] == 1;
}
public String normalize(List<int[]> cells, int[][] transformations) {
List<String> forms = new ArrayList();
for (int[] tran : transformations) {
List<int[]> list1 = new ArrayList();
List<int[]> list2 = new ArrayList();
for (int[] cell : cells) {
int x = cell[0];
int y = cell[1];
// Generate the 8 different transformations
// (x, y), (x, -y), (-x, y), (-x, -y)
// (y, x), (-y, x), (y, -x), (-y, -x)
list1.add(new int[] {x * tran[0], y * tran[1]});
list2.add(new int[] {y * tran[1], x * tran[0]});
}
}
// Take the first one the represent the shape of the island
Collections.sort(forms);
return forms.get(0);
}
public String getKey(List<int[]> cells) {
// Sort the cells and the take the first cell in sorted order as the origin
Collections.sort(cells, new Comparator<int[]>(){
@Override
public int compare(int[] a, int[] b) {
return a[0] == b[0] ? a[1] - b[1] : a[0] - b[0];
}
});
StringBuilder key = new StringBuilder();
int originX = cells.get(0)[0], originY = cells.get(0)[1];
// Generate the key
for (int[] cell : cells) {
key.append((cell[0] - originX) + ":" + (cell[1] - originY));
}
return key.toString();
}
}

## 3. Time & Space Complexity

DFS: 时间复杂度O(mn * log(mn)), 空间复杂度O(mn)