JS实现广度优先搜索得到两点间最短路径

广度优先搜索

代码：

const map = [
[0, 1, <
ad18
span class="hljs-number">1, 0, 1],
[1, 0, 0, 1, 0],
[1, 0, 0, 0, 1],
[0, 1, 0, 0, 0],
[1, 0, 1, 0, 0]
]

function bfs(map, start, end){
let queue = [],
vertex_num = map.length,    //节点数
count = 0,      //迭代次数
unvisited = [], //保存还未访问过的节点
step_dict = []; //记录步长

//init
for(let i=0; i<vertex_num; i++){
unvisited[i] = true
step_dict[i] = 0
}
queue.push(start)

while(queue.length){
let flag = false    //若遇到孤立节点，则跳过而不加count
let cur = queue.shift()
step_dict[cur]++
unvisited[cur] = false

for(let i=0; i<vertex_num; i++){
if(map[cur][i] === 1 && unvisited[i] === true){
queue.push(i)
unvisited[i] = false
step_dict[i] = count
flag = true
}

}
if(flag){
count++
}
}

return step_dict[end]
}
//测试
console.log(bfs(map,3,4))

output

3