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
思路：用string来表示移动状态，那么每个形状一样的，做dfs，那么得到的string应该是一样的。注意最后要加个“b“
因为直接down -> left,不等同于down-> deadend -> back -> left; 这是两种不同的移动方法；
class Solution {
public int numDistinctIslands(int[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int n = grid.length;
int m = grid[0].length;
HashSet<String> set = new HashSet<String>();
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(grid[i][j] == 1) {
StringBuilder sb = new StringBuilder();
dfs(grid, sb, i, j, "o");
set.add(sb.toString());
}
}
}
return set.size();
}
private int[] dx = {0,0,-1,1};
private int[] dy = {-1,1,0,0};
private void dfs(int[][] grid, StringBuilder sb, int x, int y, String direction) {
if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
return;
}
if(grid[x][y] == 1) {
grid[x][y] = 0;
sb.append(direction);
dfs(grid, sb, x + 1, y, "d");
dfs(grid, sb, x - 1, y, "u");
dfs(grid, sb, x, y + 1, "r");
dfs(grid, sb, x, y - 1, "l");
sb.append("b"); // 因为直接down -> left,不等同于down-> deadend -> back -> left;
}
}
}