原文:
给定一个，任务是找出普鲁弗序列所构成的树的所有节点的度。 例:
input: arr[] = {4, 1, 3, 4}
output: 2 1 2 3 1 1
the tree is:
2----4----3----1----5
|
6
input: arr[] = {1, 2, 2}
output: 2 3 1 1 1
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。 高效方法:创建一个大小为 2 的度【】数组，大于 prufer 序列的长度，因为如果 n 是节点数，那么 prufer 序列的长度就是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。 以下是上述方法的实施:
c
// c implementation of the approach
#include
using namespace std;
// function to print the degrees of every
// node in the tree made by
// the given prufer sequence
void printdegree(int prufer[], int n)
{
int node = 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 <= node; i )
degree[i] = 1;
// increase the count of the degree
for (int i = 0; i < n; i )
degree[prufer[i]] ;
// print the degree of every node
for (int i = 1; i <= node; i ) {
cout << degree[i] << " ";
}
}
// driver code
int main()
{
int a[] = { 4, 1, 3, 4 };
int n = sizeof(a) / sizeof(a[0]);
printdegree(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.util.*;
class gfg
{
// function to print the degrees of every
// node in the tree made by
// the given prufer sequence
static void printdegree(int prufer[], int n)
{
int node = 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 <= node; i )
{
degree[i] = 1;
}
// increase the count of the degree
for (int i = 0; i < n; i )
{
degree[prufer[i]] ;
}
// print the degree of every node
for (int i = 1; i <= node; i )
{
system.out.print(degree[i] " ");
}
}
// driver code
public static void main(string[] args)
{
int a[] = {4, 1, 3, 4};
int n = a.length;
printdegree(a, n);
}
}
/* this code contributed by princiraj1992 */
python 3
# python3 implementation of the approach
# function to print the degrees of
# every node in the tree made by
# the given prufer sequence
def printdegree(prufer, n):
node = n 2
# hash-table to mark the
# degree of every node
degree = [1] * (n 2 1)
# increase the count of the degree
for i in range(0, n):
degree[prufer[i]] = 1
# print the degree of every node
for i in range(1, node 1):
print(degree[i], end = " ")
# driver code
if __name__ == "__main__":
a = [4, 1, 3, 4]
n = len(a)
printdegree(a, n)
# this code is contributed by rituraj jain
c
// c# implementation of the approach
using system;
class gfg
{
// function to print the degrees of every
// node in the tree made by
// the given prufer sequence
static void printdegree(int []prufer, int n)
{
int node = 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 <= node; i )
{
degree[i] = 1;
}
// increase the count of the degree
for (int i = 0; i < n; i )
{
degree[prufer[i]] ;
}
// print the degree of every node
for (int i = 1; i <= node; i )
{
console.write(degree[i] " ");
}
}
// driver code
public static void main(string[] args)
{
int []a = {4, 1, 3, 4};
int n = a.length;
printdegree(a, n);
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
2 1 2 3 1 1
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