---
id: 587d825c367417b2b2512c90
title: 广度优先搜索
challengeType: 1
videoUrl: ''
dashedName: breadth-first-search
---
# --description--
到目前为止，我们已经学会了创建图表表示的不同方法。现在怎么办？一个自然的问题是图中任何两个节点之间的距离是多少？输入图遍历算法 。 遍历算法是遍历或访问图中节点的算法。一种遍历算法是广度优先搜索算法。该算法从一个节点开始，首先访问一个边缘的所有邻居，然后继续访问它们的每个邻居。在视觉上，这就是算法正在做的事情。 要实现此算法，您需要输入图形结构和要启动的节点。首先，您需要了解距起始节点的距离。这个你想要开始你所有的距离最初一些大的数字，如`Infinity` 。这为从起始节点无法访问节点的情况提供了参考。接下来，您将要从开始节点转到其邻居。这些邻居是一个边缘，此时你应该添加一个距离单位到你要跟踪的距离。最后，有助于实现广度优先搜索算法的重要数据结构是队列。这是一个数组，您可以在其中添加元素到一端并从另一端删除元素。这也称为FIFO或先进先出数据结构。
# --instructions--
编写一个函数`bfs()` ，它将邻接矩阵图（二维数组）和节点标签根作为参数。节点标签只是`0`到`n - 1`之间节点的整数值，其中`n`是图中节点的总数。您的函数将输出JavaScript对象键值对与节点及其与根的距离。如果无法到达节点，则其距离应为`Infinity` 。
# --hints--
输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` ，起始节点为`1`应该返回`{0: 1, 1: 0, 2: 1, 3: 2}`
```js
assert(
(function () {
var graph = [
[0, 1, 0, 0],
[1, 0, 1, 0],
[0, 1, 0, 1],
[0, 0, 1, 0]
];
var results = bfs(graph, 1);
return isEquivalent(results, { 0: 1, 1: 0, 2: 1, 3: 2 });
})()
);
```
输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` ，起始节点为`1`应该返回`{0: 1, 1: 0, 2: 1, 3: Infinity}`
```js
assert(
(function () {
var graph = [
[0, 1, 0, 0],
[1, 0, 1, 0],
[0, 1, 0, 0],
[0, 0, 0, 0]
];
var results = bfs(graph, 1);
return isEquivalent(results, { 0: 1, 1: 0, 2: 1, 3: Infinity });
})()
);
```
输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` ，起始节点为`0`应该返回`{0: 0, 1: 1, 2: 2, 3: 3}`
```js
assert(
(function () {
var graph = [
[0, 1, 0, 0],
[1, 0, 1, 0],
[0, 1, 0, 1],
[0, 0, 1, 0]
];
var results = bfs(graph, 0);
return isEquivalent(results, { 0: 0, 1: 1, 2: 2, 3: 3 });
})()
);
```
起始节点为`0`的输入图`[[0, 1], [1, 0]]`应返回`{0: 0, 1: 1}`
```js
assert(
(function () {
var graph = [
[0, 1],
[1, 0]
];
var results = bfs(graph, 0);
return isEquivalent(results, { 0: 0, 1: 1 });
})()
);
```
# --seed--
## --after-user-code--
```js
// Source: http://adripofjavascript.com/blog/drips/object-equality-in-javascript.html
function isEquivalent(a, b) {
// Create arrays of property names
var aProps = Object.getOwnPropertyNames(a);
var bProps = Object.getOwnPropertyNames(b);
// If number of properties is different,
// objects are not equivalent
if (aProps.length != bProps.length) {
return false;
}
for (var i = 0; i < aProps.length; i++) {
var propName = aProps[i];
// If values of same property are not equal,
// objects are not equivalent
if (a[propName] !== b[propName]) {
return false;
}
}
// If we made it this far, objects
// are considered equivalent
return true;
}
```
## --seed-contents--
```js
function bfs(graph, root) {
var nodesLen = {};
return nodesLen;
};
var exBFSGraph = [
[0, 1, 0, 0],
[1, 0, 1, 0],
[0, 1, 0, 1],
[0, 0, 1, 0]
];
console.log(bfs(exBFSGraph, 3));
```
# --solutions--
```js
function bfs(graph, root) {
var nodesLen = {};
// Set all distances to infinity
for (var i = 0; i < graph.length; i++) {
nodesLen[i] = Infinity;
}
nodesLen[root] = 0; // ...except root node
var queue = [root]; // Keep track of nodes to visit
var current; // Current node traversing
// Keep on going until no more nodes to traverse
while (queue.length !== 0) {
current = queue.shift();
// Get adjacent nodes from current node
var curConnected = graph[current]; // Get layer of edges from current
var neighborIdx = []; // List of nodes with edges
var idx = curConnected.indexOf(1); // Get first edge connection
while (idx !== -1) {
neighborIdx.push(idx); // Add to list of neighbors
idx = curConnected.indexOf(1, idx + 1); // Keep on searching
}
// Loop through neighbors and get lengths
for (var j = 0; j < neighborIdx.length; j++) {
// Increment distance for nodes traversed
if (nodesLen[neighborIdx[j]] === Infinity) {
nodesLen[neighborIdx[j]] = nodesLen[current] + 1;
queue.push(neighborIdx[j]); // Add new neighbors to queue
}
}
}
return nodesLen;
}
```