2022携程暑期实习后端开发二面面经
2022年3月25日13点30-14点10
没有自我介绍
上来直接问你对后端开发的研究有哪些？
我结合自己做的项目，从JavaWeb讲到了SpringBoot
接下来就做面试官出的算法题
public class Main {
public static void main(String[] args) {
Flight f1 = new Flight("F001", "SHA", "HKG");
Flight f2= new Flight("F002", "SHA", "TYO");
Flight f3 = new Flight("F003", "HKG", "NYC");
Flight f4 = new Flight("F004", "TYO", "NYC");
Flight f5 = new Flight("F005", "HKG", "TYO");
Flight f6 = new Flight("F006", "TYO", "HKG");
Flight f7 = new Flight("F007", "TYO", "OSA");
Flight f8 = new Flight("F008", "OSA", "NYC");
Flight f9 = new Flight("F009", "SHA", "OSA");
}
static class Flight{
String flightId;
String depart;
String arrival;
Flight (String id, String start, String end) {
this.flightId = id;
this.depart = start;
this.arrival = end;
}
}
} 以上代码是面试官聊天框里发过来的题目：要求飞机从上海飞到纽约的所有路径 我的想法是：这是一道图论算法题，采用dfs就行了
由于这是一道应用题，不像力扣牛客那样直接秒就行了，还需要处理一下数据
我一开始光讲，面试官建议我画一下，我就用画图演示了一下，表示需要将数据转换成邻接表的形式
因为Java处理字符串比较麻烦，所以我想用数字代替字符串，其实是一样的，面试官同意了
以下是我的解题代码
public class Main {
static List<List<Integer>> res=new LinkedList<>();
static LinkedList<Integer> path=new LinkedList<>();
public static void main(String[] args) {
Flight f1 = new Flight("F001", 0, 1);
Flight f2= new Flight("F002", 0, 2);
Flight f3 = new Flight("F003", 1, 4);
Flight f4 = new Flight("F004", 2, 4);
Flight f5 = new Flight("F005", 1, 2);
Flight f6 = new Flight("F006", 2, 1);
Flight f7 = new Flight("F007", 2, 3);
Flight f8 = new Flight("F008", 3, 4);
Flight f9 = new Flight("F009", 0, 3);
int[][] graph=new int[][]{{1,2,3},{2,4},{1,3,4},{3,4},{}};
boolean[] vis=new boolean[5];
dfs(graph,0,4,vis);
for(List<Integer> a:res){
for (int b:a){
System.out.print(b);
}
System.out.println();
}
}
static class Flight{
String flightId;
Integer depart;
Integer arrival;
Flight (String id, Integer start, Integer end) {
this.flightId = id;
this.depart = start;
this.arrival = end;
}
}
public static void dfs(int[][] graph,int cur,int end,boolean[] vis){
if (vis[cur]==true)return;
path.add(cur);
vis[cur]=true;
if (cur==end){
res.add(new LinkedList<>(path));
vis[cur]=false;
path.removeLast();
return;
}
for (int a:graph[cur]){
dfs(graph,a,end,vis);
}
vis[cur]=false;
path.removeLast();
}
} 最后反问 结束
#携程##实习##后端开发##面经#
查看8道真题和解析