[LeetCode] 694. Number of Distinct Islands

Problem

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.

Solution

class Solution {
    public int numDistinctIslands(int[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        Set<String> set = new HashSet<>();
        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();
                    dfs(grid, i, j, sb, "start:");
                    grid[i][j] = 0;
                    System.out.println(sb.toString());
                    set.add(sb.toString());
                }
            }
        }
        return set.size();
    }
    
    private void dfs(int[][] grid, int i, int j, StringBuilder sb, String dir) {
        if (i < 0 || i == grid.length || j < 0 || j == grid[0].length || grid[i][j] == 0) return;
        sb.append(dir);
        grid[i][j] = 0;
        dfs(grid, i-1, j, sb, "d");
        dfs(grid, i+1, j, sb, "u");
        dfs(grid, i, j-1, sb, "l");
        dfs(grid, i, j+1, sb, "r");
        sb.append("#");
    }
}
阅读 2.1k

推荐阅读
Road to Glory
用户专栏

對酒當歌,人生幾何?譬如朝露,去日苦多。

22 人关注
639 篇文章
专栏主页