原文:
给定一个树的 ，任务是打印给定 prufer 序列的树中具有最大度的节点。如果最大度数的节点很多，打印最小度数的节点。 例:
input: a[] = {4, 1, 3, 4}
output: 4
the tree is:
2----4----3----1----5
|
6
input: a[] = {1, 2, 2}
output: 2
一种简单的方法是使用普鲁夫序列创建树，然后找到所有节点的度，然后找到其中的最大值。 高效方法:创建一个大小为 2 的度【】数组，大于普鲁弗序列的长度，因为如果 n 是节点数，普鲁弗序列的长度是n–2。首先，用 1 填充度数数组。迭代普鲁弗序列，增加每个元素在度表中的出现频率。这种方法之所以有效，是因为普鲁弗序列中一个节点的频率比树中的度数少一个。现在迭代度数组，找到最大频率的节点，这就是我们的答案。 以下是上述方法的实施:
c
// c implementation of the approach
#include
using namespace std;
// function to return the node with
// the maximum degree in the tree
// whose prufer sequence is given
int findmaxdegreenode(int prufer[], int n)
{
int nodes = n 2;
// hash-table to mark the
// degree of every node
int degree[n 2 1];
// initially let all the degrees be 1
for (int i = 1; i <= nodes; i )
degree[i] = 1;
// increase the count of the degree
for (int i = 0; i < n; i )
degree[prufer[i]] ;
int maxdegree = 0;
int node = 0;
// find the node with maximum degree
for (int i = 1; i <= nodes; i ) {
if (degree[i] > maxdegree) {
maxdegree = degree[i];
node = i;
}
}
return node;
}
// driver code
int main()
{
int a[] = { 1, 2, 2 };
int n = sizeof(a) / sizeof(a[0]);
cout << findmaxdegreenode(a, n);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of the approach
import java.io.*;
class gfg
{
// function to return the node with
// the maximum degree in the tree
// whose prufer sequence is given
static int findmaxdegreenode(int prufer[], int n)
{
int nodes = n 2;
// hash-table to mark the
// degree of every node
int []degree = new int[n 2 1];
// initially let all the degrees be 1
for (int i = 1; i <= nodes; i )
degree[i] = 1;
// increase the count of the degree
for (int i = 0; i < n; i )
degree[prufer[i]] ;
int maxdegree = 0;
int node = 0;
// find the node with maximum degree
for (int i = 1; i <= nodes; i )
{
if (degree[i] > maxdegree)
{
maxdegree = degree[i];
node = i;
}
}
return node;
}
// driver code
public static void main (string[] args)
{
int []a = { 1, 2, 2 };
int n = a.length;
system.out.println(findmaxdegreenode(a, n));
}
}
// this code is contributed by ajit_00023
python 3
# python implementation of the approach
# function to return the node with
# the maximum degree in the tree
# whose prufer sequence is given
def findmaxdegreenode(prufer, n):
nodes = n 2;
# hash-table to mark the
# degree of every node
degree = [0]*(n 2 1);
# initially let all the degrees be 1
for i in range(1,nodes 1):
degree[i] = 1;
# increase the count of the degree
for i in range(n):
degree[prufer[i]] =1;
maxdegree = 0;
node = 0;
# find the node with maximum degree
for i in range(1,nodes 1):
if (degree[i] > maxdegree):
maxdegree = degree[i];
node = i;
return node;
# driver code
a = [ 1, 2, 2 ];
n = len(a);
print(findmaxdegreenode(a, n));
# this code has been contributed by 29ajaykumar
c
// c# implementation of the approach
using system;
class gfg
{
// function to return the node with
// the maximum degree in the tree
// whose prufer sequence is given
static int findmaxdegreenode(int []prufer, int n)
{
int nodes = n 2;
// hash-table to mark the
// degree of every node
int []degree = new int[n 2 1];
// initially let all the degrees be 1
for (int i = 1; i <= nodes; i )
degree[i] = 1;
// increase the count of the degree
for (int i = 0; i < n; i )
degree[prufer[i]] ;
int maxdegree = 0;
int node = 0;
// find the node with maximum degree
for (int i = 1; i <= nodes; i )
{
if (degree[i] > maxdegree)
{
maxdegree = degree[i];
node = i;
}
}
return node;
}
// driver code
static public void main ()
{
int []a = { 1, 2, 2 };
int n = a.length;
console.writeline(findmaxdegreenode(a, n));
}
}
// this code is contributed by ankitrai01
java 描述语言
output:
2
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