1004 Counting Leaves Java

1004 Counting Leaves (30 分)

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner input=new Scanner(System.in);
        int totalNode=input.nextInt();
        int innerNode=input.nextInt();
        Map<Integer,List<Integer>> nodeMap=new HashMap<>();;
        for(int i=0;i<innerNode;i++){
            int No= input.nextInt();
            nodeMap.put(No,new ArrayList<>());
            int number= input.nextInt();
            for(int j=0;j<number;j++)
                nodeMap.get(No).add(input.nextInt());
        }

        List<Integer> number=new ArrayList<Integer>();
        List<Integer> queue=new ArrayList<Integer>();
        queue.add(1);

        int cnt=0;
        int last=1;
        while(queue.size()>0){
            int temp=queue.get(0);
            if(nodeMap.get(temp)==null) {
                cnt++;
            }else {
                for(int i=0;i<nodeMap.get(temp).size();i++){
                    queue.add(nodeMap.get(temp).get(i));
                }
            }
            if(temp==last){
                    last=queue.get(queue.size()-1);
                    number.add(cnt);
                    cnt=0;
            }
            queue.remove(0);
        }


        for(int i=0;i<number.size()-1;i++){
            System.out.print(number.get(i)+" ");
        }
        System.out.print(number.get(number.size()-1));
        }
        
}

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来源： https://blog.csdn.net/weixin_45549192/article/details/113972987