递归遍历二维数组，图的邻接矩阵，深度优先搜索dfs

一.递归遍历二维数组，即深度优先搜索

　　主要是向右，向下搜索

public class DFSTest1 {

    public static void main(String[] args) {
        int[][] a1 = new int[][]{
                {1,2,3,4},
                {5,6,7,8},
                {9,10,11,12},
        };
        int row = a1.length;
        int col = a1[0].length;

        boolean[][] visited = new boolean[row][col];
        dfs(a1,0,0,visited);

    }

    private static void dfs(int[][] nums,int x,int y,boolean[][] visited){
        int row = nums.length;
        int col = nums[0].length;
        if(x>=row){
            return;
        }
        if(y>=col){
            return;
        }
        if(visited[x][y]){
            return;
        }
        System.out.print(nums[x][y] +",");
        visited[x][y] = true;

        dfs(nums,x+1,y,visited);
        dfs(nums,x,y+1,visited);
    }
}

输出: 1,5,9,10,11,12,6,7,8,2,3,4,

搜索方向如下图所示:

二.递归遍历图,基于邻接矩阵，即深度优先搜索

//无向无权图
public class Graph {
    //顶点数量
    private int vertexNum;
    //边的数量
    private int edgeNum;
    //邻接矩阵
    private int[][] adjustMatrix;

    public Graph(int v){
        vertexNum = v;
        adjustMatrix = new int[v][v];
    }

    public boolean addEdge(int begin,int end){
        if(begin<0||begin>=vertexNum||end<0||end>=vertexNum){
            return false;
        }
        adjustMatrix[begin][end] = 1;
        adjustMatrix[end][begin] = 1;
        edgeNum++;
        return true;
    }

    public void dfs(int start, Set<Integer> visited){
        System.out.println(start);
        visited.add(start);
        for (int i = 0; i < vertexNum; i++) {
            if(!visited.contains(i) && adjustMatrix[start][i]==1){
                dfs(i,visited);
            }
        }
    }
}

public class GraphTest {
    public static void main(String[] args) {
        Graph graph = new Graph(5);
        graph.addEdge(0,1);
        graph.addEdge(0,2);
        graph.addEdge(0,3);
        graph.addEdge(2,3);
        graph.addEdge(2,4);
        graph.addEdge(1,4);
        Set<Integer> visited = new HashSet<>();
        graph.dfs(0,visited);
    }
}