非递归实现广度优先遍历（BFS）

• 对于广度优先遍历，使用递归方式实现非常普遍，这里我们讨论一下如何使用非递归来实现

算法

• 1、以图的邻接矩阵为例
• 2、使用一个queue来保存访问顺序，每次取队首元素，如果其有邻接点，则将其所有邻接点入队，并设置visit数组为1，最后再将其出队
• 3、循环结束条件为：队列为空

代码实现

以下图为例

public class BfsIterateAdjacentMatrix {
    private static Map<Integer, String> idVertexMap = new HashMap<Integer, String>(){{
        put(0,"V0");
        put(1,"V1");
        put(2,"V2");
        put(3,"V3");
        put(4,"V4");
        put(5,"V5");
        put(6,"V6");
        put(7,"V7");

    }};

    private static int[][] adjacentMatrix = new int[][]{
            //V0,V1,V2,V3,V4,V5,V6,V7
            {0, 1, 1, 0, 0, 0, 0, 0},
            {1, 0, 0, 1, 1, 0, 0, 0},
            {1, 0, 0, 0, 0, 1, 1, 0},
            {0, 1, 0, 0, 0, 0, 0, 1},
            {0, 1, 0, 0, 0, 0, 0, 1},
            {0, 0, 1, 0, 0, 0, 0, 0},
            {0, 0, 1, 0, 0, 0, 0, 0},
            {0, 0, 0, 1, 1, 0, 0, 0},
    };

    private static int[] visited = new int[adjacentMatrix.length];

    public static void main(String[] args){
        bfs();
    }

    private static void bfs(){
        Queue<Integer> visitQueue = new LinkedBlockingQueue<>();
        visitQueue.add(0);
        int lineNo = 0;
        visited[lineNo] = 1;

        while (!visitQueue.isEmpty()){
            lineNo = visitQueue.peek();
            System.out.println("visit: " + idVertexMap.get(lineNo));
            
            for (int colNo =0; colNo < adjacentMatrix.length; colNo++){
                if (visited[colNo] ==0 && adjacentMatrix[lineNo][colNo]==1){
                    visitQueue.add(colNo);
                    visited[colNo]=1;
                }
            }
            visitQueue.poll();
        }
    }
}

来源：CSDN

作者：iNiegang

链接：https://blog.csdn.net/iNiegang/article/details/104117006