题目
题目链接：
https://www.nowcoder.com/practice/64a4c026b2aa4411984f560deec36323
思路
图，BFS，队列
参考答案Java
java
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param numProject int整型
* @param groups int整型ArrayList<ArrayList<>>
* @return int整型ArrayList
*/
public ArrayList<Integer> findOrder (int numProject,
ArrayList<ArrayList<Integer>> groups) {
//map表示图
Map<Integer, Gnode> graph = new HashMap<>();
for (int i = 0; i < numProject ; i++) {
graph.put(i, new Gnode(i));
}
for (ArrayList<Integer> group :
groups) { //xxxf代表 开始节点 xxxt代表 f的邻居
int vf = group.get(1);
int vt = group.get(0);
Gnode nodef = graph.get(vf);
Gnode nodet = graph.get(vt);
nodet.in++;
nodef.nexts.add(nodet);
}
Queue<Gnode> q0 = new LinkedList<>();
Set<Integer> set = new HashSet<>();
for (Integer v : graph.keySet()) {
if (graph.get(v).in == 0) {
q0.add(graph.get(v));
set.add(v);
}
}
ArrayList<Integer> ll = new ArrayList<>();
while (!q0.isEmpty()) {
int size = q0.size();
for (int i = 0; i < size ; i++) {
Gnode cur = q0.poll();
ll.add(cur.data);
for (Gnode next : cur.nexts) {
if (set.contains(next.data)) {
return new ArrayList<>();//出现环了，直接返回空的ArrayList
}
if (--next.in == 0) {
q0.add(next);
set.add(next.data);
}
}
}
}
return ll.size() == numProject ? ll : new ArrayList<>();
}
static class Gnode {
int in;
int data;
List<Gnode> nexts;
public Gnode(int d) {
data = d;
in = 0;
nexts = new ArrayList<>();
}
}
}参考答案Go
go
package main
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param numProject int整型
* @param groups int整型二维数组
* @return int整型一维数组
*/
func findOrder(numProject int, groups [][]int) []int {
// map表示图
graph := map[int]*Gnode{}
for i := 0; i < numProject; i++ {
graph[i] = &Gnode{i, 0, []*Gnode{}}
}
for _, gp := range groups {
vf := gp[1] //出发
vt := gp[0] //到达
nodef := graph[vf]
nodet := graph[vt]
nodef.nexts = append(nodef.nexts, nodet) //增加邻居
nodet.in++ //入度+1
}
q0 := []*Gnode{} //Go中队列用切片来表示
set := map[int]bool{}
for _, v1 := range graph {
if v1.in == 0 {
q0 = append(q0, v1)
set[v1.data] = true
}
}
ll := []int{}
for len(q0) > 0 {
size := len(q0)
q0bak := []*Gnode{}
for i := 0; i < size; i++ {
cur := q0[i]
//_,ok:=set[cur.data]
ll = append(ll, cur.data)
for _, next := range cur.nexts {
next.in--
if next.in == 0 {
_, ok := set[next.data]
if ok {
return []int{}
}
q0bak = append(q0bak, next)
set[next.data] = true
}
}
}
q0 = q0bak
}
if len(ll) == numProject {
return ll
}
return []int{}
}
type Gnode struct { //图的节点
data int
in int
nexts []*Gnode
}参考答案PHP
php
<?php
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param numProject int整型
* @param groups int整型二维数组
* @return int整型一维数组
*/
function findOrder( $numProject , $groups )
{
// 图用map表示,PHP中数组是万能，数组也是java中的map,set,list
$graph = array();
for($i=0;$i<$numProject;$i++){
$graph[$i] = new Gnode($i);
}
foreach ($groups as $v){
$vt= $v[0];
$vf =$v[1];
$nodet = $graph[$vt]; //出发节点
$nodef = $graph[$vf]; //到达节点，也就是出发节点的邻居
$nodet->in++;
array_push( $nodef->nexts,$nodet);
}
//BFS
$q0 = [];
$set =[];
foreach ($graph as $node){
if($node -> in ==0){
$q0[count($q0)] = $node; //放进入度为0的队列
$set[$node->data] = $node->data; //访问过该节点了
}
}
$ll = [];
while (count($q0) >0){
$size = count($q0);
$q0bak = [];
for($i=0;$i<$size;$i++){
$cur =$q0[$i];
$ll[count($ll)] = $cur->data;
foreach ($cur->nexts as $next){
$next->in--;
if($next->in ==0){
if(isset($set[$next->data])){
return []; //出现环了
}
$q0bak[count($q0bak)] = $next;
$set[$next->data] = $next->data;
}
}
}
$q0 =$q0bak;
}
if(count($ll) == $numProject){
return $ll;
}
return [];
}
class Gnode{
public $in;
public $data;
public $nexts;
public function __construct($d)
{
$this->data = $d;
$this->in =0;
$this->nexts = [];
}
}