Group of 1s in a Matrix

最新推荐文章于 2025-12-15 21:49:55 发布

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本文介绍了一种使用深度优先搜索（DFS）算法来计算矩阵中由相邻1组成的团的数量的方法。通过遍历矩阵并标记已访问的单元格，确保不重复计算同一团内的1。实例代码演示了如何实现这一过程。

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, 1, 1}, {1, 0 ,0 , 1, 0}, {1, 1, 1, 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) {
		if (i < 0 || i >= input.length || j < 0 || j >= input[0].length 
				|| visited[i][j] == true || input[i][j] == 0) return;
		visited[i][j] = true;
		traverse(input, visited, i - 1, j);
		traverse(input, visited, i + 1, j);
		traverse(input, visited, i, j - 1);
		traverse(input, visited, i, j + 1);
	}
}

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