原题链接在这里：https://leetcode.com/problems/number-of-distinct-islands/description/
题目：
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.
题解：
对每个为1的点进行dfs, 同时记下做过的path. 看最后做过多少种不同的path.
Note: dfs每层最后要加分割的数字, 不然不能确定step是在哪个level走的.
e.g. 1-1-1 和 1-1 如果不加分割数字走出的path是相同的.
| |
1 1- 1
Time Complexity: O(m*n). m = grid.length. n = grid[0].length.
Space: O(m*n). stack space.
AC Java:
1 class Solution { 2 public int numDistinctIslands(int[][] grid) { 3 if(grid == null || grid.length == 0 || grid[0].length == 0){ 4 return 0; 5 } 6 7 HashSet<List<Integer>> shapes = new HashSet<List<Integer>>(); 8 for(int i = 0; i<grid.length; i++){ 9 for(int j = 0; j<grid[0].length; j++){ 10 if(grid[i][j] == 1){ 11 List<Integer> shape = new ArrayList<Integer>(); 12 dfs(grid, i, j, shape, 0); 13 shapes.add(shape); 14 } 15 } 16 } 17 18 return shapes.size(); 19 } 20 21 private void dfs(int[][] grid, int i, int j, List<Integer> shape, int direction){ 22 if(i<0 || i>=grid.length || j<0 || j>=grid[0].length || grid[i][j]!=1){ 23 return; 24 } 25 26 grid[i][j] = 0; 27 shape.add(direction); 28 dfs(grid, i-1, j, shape, 1); 29 dfs(grid, i+1, j, shape, 2); 30 dfs(grid, i, j-1, shape, 3); 31 dfs(grid, i, j+1, shape, 4); 32 33 shape.add(0); 34 } 35 }