Group of 1s in a Matrix

Given a matrix with 1s and 0s, please find the number of groups of 1s. A group is defined by horizontally or vertically adjacent 1s. For example, there are four groups of 1s in figure below.

Analysis: Use DFS.

public class NumberOfGroups {
	
	public static void main(String[] args) {
		int[][] input = {{1, 1, 0, 0, 1}, {1, 0 ,0 , 1, 0}, {1, 1, 0, 1, 0}, {0, 0, 1, 0, 0}};
		System.out.println(new NumberOfGroups().getNumberofGroups(input));
	}
	
	public int getNumberofGroups(int[][] input) {
		boolean[][] visited = new boolean[input.length][input[0].length];
		int count = 0;
		for (int i = 0; i < input.length; i++) {
			for (int j = 0; j < input[0].length; j++) {
				if (input[i][j] == 1 && visited[i][j] == false) {
					count++;
					traverse(input, visited, i, j);
				}
			}
		}
		return count;
	}
	
	public void traverse(int[][] input, boolean[][] visited, int i, int j) {
		visited[i][j] = true;
		if (i - 1 >= 0 && visited[i - 1][j] == false && input[i - 1][j] == 1) {
			traverse(input, visited, i - 1, j);
		}
		if (i + 1 < input.length && visited[i + 1][j] == false && input[i + 1][j] == 1) {
			traverse(input, visited, i + 1, j);
		}
		if (j - 1 >= 0 && visited[i][j - 1] == false && input[i][j - 1] == 1) {
			traverse(input, visited, i, j - 1);
		}
		if (j + 1 < input[0].length && visited[i][j + 1] == false && input[i][j + 1] == 1) {
			traverse(input, visited, i, j + 1);
		}
	}
}