问题描述
我正在研究 BFS 算法,但很难弄清楚如何跟踪最短路径。
在我使用的代码下方:
const graph = {
1: [2,3,4],2: [5,6],3: [10],4: [7,8],5: [9,10],7: [11,12],11: [13],};
function bfs(graph,start,end) {
let queue = [...graph[start]];
let path = [start];
let searched = [];
while (queue.length > 0) {
let curVert = queue.shift();
if (curVert === end) {
return path;
} else if (searched.indexOf(curVert) === -1 && graph[curVert]) {
queue = [...queue,...graph[curVert]];
searched.push(curVert);
path.push(curVert);
}
}
}
console.log(bfs(graph,1,13));
作为函数调用的回报,我想得到的是最短路径。在本例中为 [1,4,7,11,13]
。
解决方法
您还需要存储每个访问过的节点的路径。
const graph = { 1: [2,3,4],2: [5,6],3: [10],4: [7,8],5: [9,10],7: [11,12],11: [13] };
function bfs(graph,start,end) {
let queue = [[start,[]]],seen = new Set;
while (queue.length) {
let [curVert,[...path]] = queue.shift();
path.push(curVert);
if (curVert === end) return path;
if (!seen.has(curVert) && graph[curVert]) {
queue.push(...graph[curVert].map(v => [v,path]));
}
seen.add(curVert);
}
}
console.log(bfs(graph,1,13));
对于每个顶点,保持其“前一个”顶点。由于没有回溯,一旦设置,前一个顶点就不会改变。同时,“上一个”地图将跟踪已经访问过的顶点。找到结束顶点后,向后迭代“上一个”映射以计算路径。
const graph = {
1: [2,11: [13],};
function bfs(graph,end) {
let queue = [start]
let prev = {[start]: null}
while (queue.length > 0) {
let curr = queue.shift();
if (curr === end) {
let path = [];
while (curr) {
path.unshift(curr);
curr = prev[curr];
}
return path;
}
if (curr in graph) {
for (let v of graph[curr]) {
if (!(v in prev)) {
prev[v] = curr;
queue.push(v);
}
}
}
}
}
console.log(bfs(graph,13));