Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input: grid = [ [“1”,“1”,“1”,“1”,“0”], [“1”,“1”,“0”,“1”,“0”], [“1”,“1”,“0”,“0”,“0”], [“0”,“0”,“0”,“0”,“0”] ] Output: 1
Input: grid = [ [“1”,“1”,“0”,“0”,“0”], [“1”,“1”,“0”,“0”,“0”], [“0”,“0”,“1”,“0”,“0”], [“0”,“0”,“0”,“1”,“1”] ] Output: 3
class GNode {
int row;
int col;
GNode(int i, int j) {
this.row = i;
this.col = j;
}
}
public int numIslands(char[][] grid) {
int[][] visited = new int[grid.length][grid[0].length];
int count = 0;
Queue<GNode> qu = new LinkedList<GNode>();
System.out.println("grid : " + grid.length);
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
System.out.println("i : " + i + " j: " + j + "vis:" + visited[i][j]);
if (grid[i][j] == '1' && visited[i][j] == 0) {
qu.add(new GNode(i, j));
bfs(i, j, grid, visited, qu);
count++;
System.out.println("count : " + count);
}
System.out.println("countaaa : " + count);
}
}
return count;
}
public void bfs(int row, int col, char[][] grid, int[][] visited, Queue<GNode> qu) {
while (!qu.isEmpty()) {
GNode current = qu.poll();
int[] xPath = new int[] { -1, 1, 0, 0 };
int[] yPath = new int[] { 0, 0, -1, 1 };
for (int x = 0; x < xPath.length; x++) {
int newX = current.row + xPath[x];
int newY = current.col + yPath[x];
if (newX >= 0 && newY >= 0
&& newX < grid.length && newY < grid[0].length
&& visited[newX][newY] == 0
&& grid[newX][newY] == '1') {
visited[newX][newY] = 1;
qu.add(new GNode(newX, newY));
}
}
}
}