广度优先搜索与深度优先搜索的 java 实现

1.广度优先搜索

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

public class Test {

static int [] distance;

//有向图的邻接矩阵表示法
static int [] [] matrix = {
{1,1,0,1,0,0,0},
{0,1,0,1,1,0,0},
{1,0,1,0,0,1,0},
{0,0,1,1,1,1,1},
{0,0,0,0,1,0,1},
{0,0,0,0,0,1,0},
{0,0,0,0,0,1,1} };

static void print( Queue<Integer> queue ) {
System.out.println(Arrays.toString(queue.toArray()));
}

//广度优先搜索
static void bfs(int s) {
Queue<Integer> q = new LinkedList<>();

//起点进队
q.offer(s);
distance[s] = 0;

print(q);

//只要队列不为空，而且 v 可到 i ，而且 i 点从未访问（很关键，用 distance[i] == -1 表示）
//distance[i] 就等于 distance[v] + 1
//并使 i 进队，每一步的队列变化将在输出中显示
while( !q.isEmpty() ) {
int v = q.poll();

for( int i = 0; i < 7; i++)
if( matrix[v][i] == 1 && distance[i] == -1 && i != v) {
distance[i] = distance[v] + 1;
q.offer(i);
}

print(q);
}
}

public static void main( String [] args) {
//初始化距离均为-1
distance = new int [7];
for( int i = 0; i < 7; i++)
distance[i] = -1;

//测试点v2
bfs(2);
System.out.println("该点到各点最短距离：" + Arrays.toString(distance));
}
}

[2]
[0, 5]
[5, 1, 3]
[1, 3]
[3, 4]
[4, 6]
[6]
[]
该点到各点最短距离：[1, 2, 0, 2, 3, 1, 3]

2.深度优先搜索

void dfs(Vertice v) {
v.visit = true;
for each Vertice w adjacent to v
if ( !w.visit )
dfs(w);
}

3.分析