694. Number of Distinct Islands
2021/4/8 10:13:18
本文主要是介绍694. Number of Distinct Islands,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
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.
1 class Solution { 2 int[][] dirs = new int[][] { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } }; 3 public int numDistinctIslands(int[][] grid) { 4 Set<String> set = new HashSet<>(); 5 int res = 0; 6 7 for (int i = 0; i < grid.length; i++) { 8 for (int j = 0; j < grid[0].length; j++) { 9 if (grid[i][j] == 1) { 10 StringBuilder sb = new StringBuilder(); 11 helper(grid, i, j, 0, 0, sb); 12 String s = sb.toString(); 13 if (!set.contains(s)) { 14 res++; 15 set.add(s); 16 } 17 } 18 } 19 } 20 return res; 21 } 22 23 public void helper(int[][] grid, int i, int j, int xpos, int ypos, StringBuilder sb) { 24 grid[i][j] = 0; 25 sb.append(xpos + "" + ypos); 26 for (int[] dir : dirs) { 27 int x = i + dir[0]; 28 int y = j + dir[1]; 29 if (x < 0 || y < 0 || x >= grid.length || y >= grid[0].length || grid[x][y] == 0) 30 continue; 31 helper(grid, x, y, xpos + dir[0], ypos + dir[1], sb); 32 } 33 } 34 }
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