Given a 2D grid
consists of 0s
(land) and 1s
(water). An island is a maximal 4-directionally connected group of 0s
and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.
Return the number of closed islands.
Example 1:
Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation:
Islands in gray are closed because they are completely surrounded by water (group of 1s).
思路:先从四个边的0开始往里面floodfill,变成1,这样剩下的,就是岛屿,再count connect component,也可以用fill函数,把0 变成1;
class Solution {
public int closedIsland(int[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int n = grid.length;
int m = grid[0].length;
// from edge into grid, change all 0 to 1;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(i == 0 || i == n - 1 || j == 0 || j == m - 1) {
if(grid[i][j] == 0) {
fill(grid, i, j);
}
}
}
}
// count connected 0 in grid;
int count = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(grid[i][j] == 0) {
count++;
fill(grid, i, j);
}
}
}
return count;
}
private int[] dx = {0,0,-1,1};
private int[] dy = {-1,1,0,0};
private void fill(int[][] grid, int x, int y) {
if(x < 0|| x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == 1) {
return;
}
grid[x][y] = 1;
for(int k = 0; k < 4; k++) {
int nx = x + dx[k];
int ny = y + dy[k];
fill(grid, nx, ny);
}
}
}