使用邻接矩阵的图的DFS（Java）

Question

Hello I'm having an issue with my code displaying the wrong order when I perform a Depth-First Traversal

class graphs{

public static void main(String [] args){
    int[][] adjMatrix = { {0, 1, 0, 0, 1, 1, 0, 0},
            {1, 0, 0, 0, 0, 1, 1, 0},
            {0, 0, 0, 1, 0, 0, 1, 0},
            {0, 0, 1, 0, 0, 0, 0, 1},
            {1, 0, 0, 0, 0, 1, 0, 0},
            {1, 1, 0, 0, 1, 0, 0, 0},
            {0, 1, 1, 0, 0, 0, 0, 1},
            {0, 0, 0, 1, 0, 0, 1, 0}};
    boolean[] visited = {false, false, false, false, false, false, false, false};
    int n = 8;

    DFS(adjMatrix, visited, n, 0);
}

public static void DFS(int[][] adjMatrix, boolean [] visited,int n, int i){
    System.out.print(" " + (i+1));
    visited[i]= true;
    for (int j = 0; j<n;j++){
        if(!(visited[j]) && adjMatrix[i][j]==1){
            DFS(adjMatrix, visited, n, j);
        }

    }
  }
}

I was told that I should be getting : 1 2 6 7 4 3 5 8

I, however, keep getting : 1 2 6 5 7 3 4 8

What am I doing wrong?

Edit: It's supposed to display the order visited. Maybe that's where I'm messing up?

Edit: Additionally, how could I have it display the order of the dead ends?

For the dead ends would something like this work:

public static void DFS(int[][] adjMatrix, boolean [] visited,int n, int i){
System.out.print(" " + (i+1));
visited[i]= true;
for (int j = 0; j<n;j++){
    if(!(visited[j]) && adjMatrix[i][j]==1){
        DFS(adjMatrix, visited, n, j);
    }

}
a.add(i+1); //assume a is an integer ArrayList and will be printed out later
}

Answer 1

You were told wrong, your output is correct.

You can check it by hand:

       1  2  3  4  5  6  7  8
    1 {0, 1, 0, 0, 1, 1, 0, 0}
    2 {1, 0, 0, 0, 0, 1, 1, 0}
    3 {0, 0, 0, 1, 0, 0, 1, 0}
    4 {0, 0, 1, 0, 0, 0, 0, 1}
    5 {1, 0, 0, 0, 0, 1, 0, 0}
    6 {1, 1, 0, 0, 1, 0, 0, 0}
    7 {0, 1, 1, 0, 0, 0, 0, 1}
    8 {0, 0, 0, 1, 0, 0, 1, 0}

You just "jump" until you find a 0, we start at 1, the first 1 in the first row is at column marked 2, we move to row 2, the first 1 in row 2 that was not yet visited/is not being visited currently is at 6, we move to row 6, again the first interesting 1 is at 5, we move to row 5, there are no interesting 1s in that row because we are currently on a path that already visited 1 and 6 so we backtrack, at this point we have 1,2,6,5. We backtrack back to row 2, first interesting 1 is at 7, in row 7 we get 3, at 3 we go to 4, then backtrack to 7 and go to 8 which finishes all nodes with your result.

@Edit: just to make it clear other answers might be possible with DFS (depending in which order you will be processing neighbours) but the answer you were supposed to provide is definitely not possible for this input.

@Edit: so to be precise it would be 1 -> 2 -> 6 -> 5 -> backtrack(2) -> 7 -> 3 -> 4 -> 8 -> end

Answer 2

根据您的矩阵，adjMatrix [6] [5] = 1但adjMatrix [6] [7] = 0，因此您的预期答案1 2 6 7 4 3 5 8不正确，因为6和7之间没有链接。应该再次检查输入数据，很有可能您输入错误。