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.
這道題和LC200 Number of Islands一樣用DFS方法妆够，重點(diǎn)在于如何記錄distinct islands shapes。
- 可以用字符串，記錄相對(duì)坐標(biāo)串。
- 可以用Integer磷籍，記錄相對(duì)坐標(biāo)串踊挠。
- 可以記錄path路徑，注意用0記錄進(jìn)出dfs函數(shù)的位置用以區(qū)分不同形狀锨苏。
比賽本應(yīng)該做出來(lái)耿币，速度慢梳杏，也沒想出記錄存儲(chǔ)的好方法。
class Solution {
ArrayList shape = new ArrayList<Integer>();
public int numDistinctIslands(int[][] grid) {
int n = grid.length;
if(n==0) return 0;
int m = grid[0].length;
Set shapes = new HashSet<ArrayList<Integer>>();
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(grid[i][j]==1){
shape = new ArrayList<Integer>();
markAsZero(grid,i,j,0);
if(!shape.isEmpty()) shapes.add(shape);
}
}
}
return shapes.size();
}
public void markAsZero(int[][] grid, int i, int j, int path){
if(i<0||j<0||i>=grid.length||j>=grid[0].length||grid[i][j]==0) return;
grid[i][j]= 0;
shape.add(path);
markAsZero(grid,i-1,j,1);
markAsZero(grid,i,j-1,2);
markAsZero(grid,i+1,j,3);
markAsZero(grid,i,j+1,4);
shape.add(0);
return;
}
}
