LeetCode 200. Number of Islands.java

Given a 2d grid map of '1's (land) and '0's (water), count 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:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

	class Solution {
	public int numIslands(char[][] grid) {
	// corner cases
	if (grid == null || grid.length == 0 || grid[0] == null || grid[0].length == 0) {
	return 0;
	}
	// main logic - 2 for loops and using every non visited land as entry point to call dfs
	int counter = 0;
	m = grid.length;
	n = grid[0].length;
	visited = new boolean[m][n];
	for (int i = 0; i < m; i++) {
	for (int j = 0; j < n; j++) {
	if ( !visited[i][j] && (grid[i][j] == '1')) {
	dfs(grid, i, j);
	counter++;
	}
	}
	}
	return counter;
	}
	private int m = 0;
	private int n = 0;
	private boolean[][] visited = null;
	private int[] dx = new int[]{0, 0, 1, -1};
	private int[] dy = new int[]{1, -1, 0, 0,};
	private void dfs(char[][] grid, int i , int j) {
	// corner cases
	if (visited[i][j] || grid[i][j] == '0') {
	return;
	}
	visited[i][j] = true;
	for (int t = 0; t < 4; t++) {
	int x = dx[t] + i;
	int y = dy[t] + j;
	if (isValid(x, y) && !visited[x][y] && (grid[x][y] == '1')) {
	dfs(grid, x, y);
	}
	}
	}
	private boolean isValid(int x, int y) {
	return x >= 0 && x < m && y >= 0 && y < n;
	}
	}

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