Problem

Given a non-empty 2D arraygridof 0's and 1's, anislandis 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 ofdistinctislands. 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.

两种思路都需要注意，不能够进行backtrack，否则只会记录一条路径

思路一（用字符串来记录x和y的坐标偏移量）

    int[][] dirs= new int[][]{{1,0},{0,1},{-1,0},{0,-1}};
    public int numDistinctIslands(int[][] grid) {
         Set<String> set= new HashSet<>();
        int res=0;

        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
                if(grid[i][j]==1) {
                    StringBuilder sb= new StringBuilder();
                    helper(grid,i,j,0,0, sb);
                    String s=sb.toString();
                    if(!set.contains(s)){
                    res++;
                    set.add(s);
}
                }
            }
        }
            return res;
    }

    public  void helper(int[][] grid,int i,int j, StringBuilder sb){
        grid[i][j]=0;
        sb.append(xpos+""+ypos);
        for(int[] dir : dirs){
            int x=i+dir[0];
            int y=j+dir[1];
            if(x<0 || y<0 || x>=grid.length || y>=grid[0].length || grid[x][y]==0) continue;
            helper(grid,x,y,xpos+dir[0],ypos+dir[1],sb);
        }
    }

思路二（用字符串来记录x和y移动的方向）

• 注意： 需要用字符记录下backtrack，虽然记录了方向，但是不知道在哪里拐弯转向其他路径。

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

        Set<String> set = new HashSet();

        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == 1) {
                    StringBuilder sb = new StringBuilder();
                    dfs(grid, i, j, 'o', sb);
                    set.add(sb.toString());
                }
            }
        }
        return set.size();
    }

    private void dfs(int[][] grid, int x, int y, char dir, StringBuilder sb) {
        if (x < 0 || y < 0 || x >= grid.length || y >= grid[0].length || grid[x][y] == 0) return;

        grid[x][y] = 0;

        sb.append(dir);

        dfs(grid, x + 1, y, 'd', sb);
        dfs(grid, x, y + 1, 'r', sb);
        dfs(grid, x - 1, y, 'u', sb);
        dfs(grid, x, y - 1, 'l', sb);

        sb.append('b');
    }