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.

这道题和LC200 Number of Islands一样用DFS方法，重点在于如何记录distinct islands shapes。

1. 可以用字符串，记录相对坐标串。
2. 可以用Integer，记录相对坐标串。
3. 可以记录path路径，注意用0记录进出dfs函数的位置用以区分不同形状。

比赛本应该做出来，速度慢，也没想出记录存储的好方法。

class Solution {
    ArrayList shape = new ArrayList<Integer>();
    public int numDistinctIslands(int[][] grid) {
        
        int n = grid.length;
        if(n==0) return 0;
        int m = grid[0].length;
        Set shapes = new HashSet<ArrayList<Integer>>();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1){
                    shape = new ArrayList<Integer>();
                    markAsZero(grid,i,j,0);
                    if(!shape.isEmpty()) shapes.add(shape);
                }
            }
        }
        return shapes.size();
    }
    
    public void markAsZero(int[][] grid, int i, int j, int path){
        if(i<0||j<0||i>=grid.length||j>=grid[0].length||grid[i][j]==0) return;
        
        grid[i][j]= 0;
        shape.add(path);
        markAsZero(grid,i-1,j,1);
        markAsZero(grid,i,j-1,2);
        markAsZero(grid,i+1,j,3);
        markAsZero(grid,i,j+1,4);
        shape.add(0);
        return;
    }
}