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1905. Count Sub Islands
You are given two m x n binary matrices grid1 and grid2 containing only 0's (representing water) and 1's (representing land). An island is a group of 1’s connected 4-directionally (horizontal or vertical). Any cells outside of the grid are considered water cells.
An island in grid2 is considered a sub-island if there is an island in grid1 that contains all the cells that make up this island in grid2.
Return the number of islands in grid2 that are considered sub-islands.
Example 1:
Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
Output: 3
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.Example 2:
Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
Output: 2
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.Constraints:
-
m == grid1.length == grid2.length -
n == grid1[i].length == grid2[i].length -
1 ⇐ m, n ⇐ 500 -
grid1[i][j]andgrid2[i][j]are either0or1.
思路分析
-
一刷（沉岛）
-
一刷（自己想法）
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/**
* 借鉴沉岛思路
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-12 14:51:03
*/
public int countSubIslands(int[][] grid1, int[][] grid2) {
if (grid1 == null || grid1.length == 0 || grid1[0].length == 0
|| grid2 == null || grid2.length == 0 || grid2[0].length == 0) {
return 0;
}
int m = grid1.length, n = grid1[0].length;
// 把不是子岛的岛屿全部沉掉
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
if (grid1[r][c] == 0 && grid2[r][c] == 1) {
dfs(grid2, r, c);
}
}
}
// 剩余岛屿即使岛屿数量
int result = 0;
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
if (grid1[r][c] == 1 && grid2[r][c] == 1) {
result++;
dfs(grid2, r, c);
}
}
}
return result;
}
private void dfs(int[][] grid, int r, int c) {
if (r < 0 || grid.length <= r
|| c < 0 || grid[r].length <= c) {
return;
}
if (grid[r][c] != 1) {
return;
}
grid[r][c] = 2;
dfs(grid, r - 1, c); // 上
dfs(grid, r + 1, c); // 下
dfs(grid, r, c - 1); // 左
dfs(grid, r, c + 1); // 右
}
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/**
* 直接判断，把不是子岛的地方修改掉，
* 因为有重复访问，所以效率更低，不如沉岛来的快
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-12 14:51:03
*/
public int countSubIslands(int[][] grid1, int[][] grid2) {
if (grid1 == null || grid1.length == 0 || grid1[0].length == 0
|| grid2 == null || grid2.length == 0 || grid2[0].length == 0) {
return 0;
}
int m = grid1.length, n = grid1[0].length;
// 剩余岛屿即使岛屿数量
int result = 0;
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
if (grid1[r][c] == 1 && grid2[r][c] == 1) {
boolean subIsland = dfs(grid1, grid2, r, c);
if (subIsland) {
result++;
}
}
}
}
return result;
}
private boolean dfs(int[][] grid1, int[][] grid2, int r, int c) {
if (r < 0 || grid1.length <= r
|| c < 0 || grid1[r].length <= c) {
return true;
}
if (grid1[r][c] == 0 && grid2[r][c] == 0) {
return true;
}
if (grid1[r][c] == 0 && grid2[r][c] == 1) {
dfs(grid2, r, c);
return false;
}
if (grid1[r][c] == 0 && grid2[r][c] == 2) {
return false;
}
if (grid1[r][c] == 1 && grid2[r][c] == 0) {
return true;
}
if (grid1[r][c] == 1 && grid2[r][c] == 2) {
return true;
}
grid2[r][c] = 2;
return dfs(grid1, grid2, r - 1, c) // 上
&& dfs(grid1, grid2, r + 1, c) // 下
&& dfs(grid1, grid2, r, c - 1) // 左
&& dfs(grid1, grid2, r, c + 1); // 右
}
private void dfs(int[][] grid, int r, int c) {
if (r < 0 || grid.length <= r
|| c < 0 || grid[r].length <= c) {
return;
}
if (grid[r][c] == 0 || grid[r][c] == 3) {
return;
}
grid[r][c] = 3;
dfs(grid, r - 1, c); // 上
dfs(grid, r + 1, c); // 下
dfs(grid, r, c - 1); // 左
dfs(grid, r, c + 1); // 右
}