694 Number of Distinct Islands

694. Number of Distinct Islands

1. 题目：

Given a non-empty 2D arraygridof 0's and 1's, an island is a group of1'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.

2. 思路

这题的难点在于如何表示island形状的唯一性，这里我们的方法是用一个List of List存储每个值为1的点与起始点的相对距离 {row, col}， 然后再把这个List of List放到Hashset里判断唯一性， 最后得到的Hashset的size就是我们要求的不同形状的岛屿的个数。时间复杂度和空间复杂度都为O(mn)

3. Implementation

class Solution {
    public int numDistinctIslands(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int m = grid.length, n = grid[0].length;
        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        Set<List<List<Integer>>> set = new HashSet<>();

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    List<List<Integer>> hashKey = new ArrayList<>();
                    searchByBFS(i, j, directions, grid, hashKey);
                    set.add(hashKey);
                }
            }
        }
        return set.size();
    }

    public void searchByBFS(int row, int col, int[][] directions, int[][] grid, List<List<Integer>> hashKey) {
        grid[row][col] = 0;
        Queue<Integer> queue = new LinkedList<>();
        int n = grid[0].length;

        queue.add(row * n + col);

        while (!queue.isEmpty()) {
            int curPoint = queue.remove();
            int curRow = curPoint / n;
            int curCol = curPoint % n;

            for (int i = 0; i < 4; i++) {
                int nextRow = curRow + directions[i][0];
                int nextCol = curCol + directions[i][1];

                if (isValid(nextRow, nextCol, grid)) {
                    grid[nextRow][nextCol] = 0;
                    hashKey.add(Arrays.asList(nextRow - row, nextCol - col));
                    queue.add(nextRow * n + nextCol);
                }
            }
        }
    } 

    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;
    }
}