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.
题目
此题是[leetcode]200. Number of Islands岛屿个数小岛题的变种
要数出不同小岛的个数，题意定义了不同小岛： 互相不能通过非反射、旋转而转化
思路
- In 2D array, once we find 1st ‘1‘, we find an island. Meanwhile, we set a ‘draw_begin‘ , ready to draw such island‘s shape
- DFS to loop through 4 different directions, checking whether another ‘1‘ is connected with. Meanwhile, we use u, d, l, r to draw island‘s shape.
- Once finish DFS, we set a ‘draw_end‘, indicating such island‘s drawing is done
- Why we use ‘draw_begin‘ and ‘draw_end‘ to set a bound ?
- 要去handle 类似以下这种情况
11 and 1 1 1 1
- Visited spots won‘t be visited again because they are updated to ‘0‘
代码
1 class Solution {
2 public int numDistinctIslands(int[][] grid) {
3 //put different drawing shape represented by string to hashset
4 Set<String> set = new HashSet<>();
5 for(int i = 0; i < grid.length; i++) {
6 for(int j = 0; j < grid[i].length; j++) {
7 if(grid[i][j] != 0) {
8 StringBuilder sb = new StringBuilder();
9 dfs(grid, i, j, sb, "draw_begin"); // set a bound
10 set.add(sb.toString());
11 }
12 }
13 }
14 // how many diffent shapes represented by string in hashset
15 return set.size();
16 }
17 private void dfs(int[][] grid, int i, int j, StringBuilder sb, String dir) {
18 // out of bound or not an island
19 if(i < 0 || i == grid.length || j < 0 || j == grid[i].length
20 || grid[i][j] == 0) return;
21
22 sb.append(dir);
23 grid[i][j] = 0;// mark a visted spot
24 dfs(grid, i-1, j, sb, "u");// up
25 dfs(grid, i+1, j, sb, "d");// down
26 dfs(grid, i, j-1, sb, "l");// left
27 dfs(grid, i, j+1, sb, "r");// right
28 sb.append("draw_end"); // set a bound
29 }
30 }
原文地址：https://www.cnblogs.com/liuliu5151/p/9825367.html