694. Number of Distinct Islands - Medium

Posted 2021-01-23 fatttcat

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.

 

hashtable + dfs

在dfs找完整islands时，用StringBuilder存island增长的path，来表示其形状（从遍历到的第一个为1的点开始，每一个点向上/下/左/右走）

比如说 11 的形状是00 10 01，  1 的的形状是00 10 1-1

          1　　　　　　　　　　11

每找到一个就存进set，防止重复。最后返回set的大小。

可以不用visited数组，直接将访问过的1赋值为0即可

时间：O(M*N)，空间：O(M*N)

class Solution {
    int[][] dirs = new int[][] {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    
    public int numDistinctIslands(int[][] grid) {
        if(grid == null || grid.length == 0) return 0;
        int m = grid.length, n = grid[0].length;
        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 sb = new StringBuilder();
                    dfs(grid, i, j, 0, 0, sb);
                    if(!set.contains(sb.toString()))
                        set.add(sb.toString());
                }
            }
        }
        return set.size();
    }
    
    public void dfs(int[][] grid, int i, int j, int xpos, int ypos, StringBuilder sb) {
        grid[i][j] = 0;
        sb.append(xpos + "" + ypos);
        for(int[] dir : dirs) {
            int x = i + dir[0], y = j + dir[1];
            if(x < 0 || x > grid.length - 1 || y < 0 || y > grid[0].length - 1 || grid[x][y] == 0)
                continue;
            dfs(grid, x, y, xpos + dir[0], ypos + dir[1], sb);
        }
    }
}