694. Number of Distinct Islands

Description

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.

Problem URL

Solution

给一二维数组表示1为岛屿，找到不同形状的岛屿，旋转相同也不可以的。是一个典型的dfs题目，双for循环遍历，如果是1（很关键）则开始dfs操作，在dfs中，如果越界或者不为1或者visited就返回，然后对四个方向进行dfs操作，最后设置对正常的返回值（回退情况）。模版题。

Using a string to denote the shape of an island. If the island have same shape, the dfs traversal order would be same order. Remember add ‘b’ for back track.

Code

class Solution {
    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] != 0){
                    StringBuilder sb = new StringBuilder();
                    dfs(grid, sb, i, j, 'o');
                    set.add(sb.toString());  
                }
            }
        }
        return set.size();
    }
    
    private void dfs(int[][] grid, StringBuilder sb, int i, int j, char dir){
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j] == 0){
            return;
        }
        sb.append(dir);
        grid[i][j] = 0;
        dfs(grid, sb, i + 1, j, 'd');
        dfs(grid, sb, i - 1, j, 'u');
        dfs(grid, sb, i, j + 1, 'r');
        dfs(grid, sb, i, j - 1, 'l');
        sb.append('b');
    }
}

Time Complexity: O(mn)
Space Complexity: O(mn)