694 Number of Distinct Islands

694. Number of Distinct Islands

1. Question

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, return1.

Example 2:

11011
10000
00001
11011

Given the above grid map, return3.

Notice that:

11
1

and

 1
11

are considered different island shapes, because we do not consider reflection / rotation.

2. Implementation

(1) DFS

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}};
        String[] symbols = {"d", "u", "l", "r"};

        Set<String> set = new HashSet<>();

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    StringBuilder key = new StringBuilder();
                    searchDistinctIslandByDFS(i, j, "s", key, directions, symbols, grid);

                    set.add(key.toString());
                }
            }
        }
        return set.size();
    }

    public void searchDistinctIslandByDFS(int row, int col, String symbol, StringBuilder key, int[][] directions, String[] symbols, int[][] grid) {
        grid[row][col] = 0;
        key.append(symbol);

        for (int i = 0; i < directions.length; i++) {
            int nextRow = row + directions[i][0];
            int nextCol = col + directions[i][1];

            if (isValid(nextRow, nextCol, grid)) {
                searchDistinctIslandByDFS(nextRow, nextCol, symbols[i], key, directions, symbols, grid);
            }
        }
        // back
        key.append("b");
    }

    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;
    }
}

3. Time & Space Complexity

DFS: 时间复杂度O(mn), 空间复杂度O(mn)