694. Number of Distinct Islands 694.独立岛屿数

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.

前提：必须岛是1才能进行后续操作

if (grid[i][j] == 1)

baseX baseY以及x y都应该是i j,然后x y再去扩展

dfs(grid, i, j, i, j, set);

每一个点都需要开新的set，不然有重复

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

class Solution {
    public int numDistinctIslands(int[][] grid) {
        //cc
        if (grid == null || grid.length == 0 || grid[0].length == 0) 
            return 0;
        
        Set<String> result = new HashSet<>();
        
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    //每一个点都需要开新的set，不然有重复
                    Set<String> set = new HashSet<>();
                    //baseX baseY以及x y都应该是i j,然后x y再去扩展
                    dfs(grid, i, j, i, j, set);
                    result.add(set.toString());
                }
            }
        }
        
        return result.size();
    }
    
    public int dfs(int[][] grid, int i, int j, int baseX, int baseY, 
                  Set<String> set) {
        //cc
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length)
            return 0;
        
        if (grid[i][j] == 1) { //前提：必须岛是1才能进行后续操作
            String distanceString = String.valueOf(i - baseX) + "_" + String.valueOf(j - baseY);
            set.add(distanceString);
            
            grid[i][j] = 0;
            dfs(grid, i + 1, j, baseX, baseY, set);
            dfs(grid, i - 1, j, baseX, baseY, set);
            dfs(grid, i, j + 1, baseX, baseY, set);
            dfs(grid, i, j - 1, baseX, baseY, set);
        }
        
        return set.size();
    }
}