给定一个具有N 个节点的无向图，任务是打印具有最小和最大度数的节点。
例子：
Input:
1-----2
| |
3-----4
Output:
Nodes with maximum degree : 1 2 3 4
Nodes with minimum degree : 1 2 3 4
Every node has a degree of 2.
Input:
1
/ \
2 3
/
4
Output:
Nodes with maximum degree : 1 2
Nodes with minimum degree : 3 4
方法：对于无向图，一个节点的度数是与它相关的边的数量，因此可以通过计算每个节点在边列表中的频率来计算每个节点的度数。因此，该方法是使用地图来计算边列表中每个顶点的频率，并使用该地图找到具有最大和最小度数的节点。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the nodes having
// maximum and minimum degree
void minMax(int edges[][2], int len, int n)
{
// Map to store the degrees of every node
map m;
for (int i = 0; i < len; i++) {
// Storing the degree for each node
m[edges[i][0]]++;
m[edges[i][1]]++;
}
// maxi and mini variables to store
// the maximum and minimum degree
int maxi = 0;
int mini = n;
for (int i = 1; i <= n; i++) {
maxi = max(maxi, m[i]);
mini = min(mini, m[i]);
}
// Printing all the nodes with maximum degree
cout << "Nodes with maximum degree : ";
for (int i = 1; i <= n; i++) {
if (m[i] == maxi)
cout << i << " ";
}
cout << endl;
// Printing all the nodes with minimum degree
cout << "Nodes with minimum degree : ";
for (int i = 1; i <= n; i++) {
if (m[i] == mini)
cout << i << " ";
}
}
// Driver code
int main()
{
// Count of nodes and edges
int n = 4, m = 6;
// The edge list
int edges[][2] = { { 1, 2 },
{ 1, 3 },
{ 1, 4 },
{ 2, 3 },
{ 2, 4 },
{ 3, 4 } };
minMax(edges, m, 4);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to print the nodes having
// maximum and minimum degree
static void minMax(int edges[][], int len, int n)
{
// Map to store the degrees of every node
HashMap m = new HashMap();
for (int i = 0; i < len; i++)
{
// Storing the degree for each node
if(m.containsKey(edges[i][0]))
{
m.put(edges[i][0], m.get(edges[i][0]) + 1);
}
else
{
m.put(edges[i][0], 1);
}
if(m.containsKey(edges[i][1]))
{
m.put(edges[i][1], m.get(edges[i][1]) + 1);
}
else
{
m.put(edges[i][1], 1);
}
}
// maxi and mini variables to store
// the maximum and minimum degree
int maxi = 0;
int mini = n;
for (int i = 1; i <= n; i++)
{
maxi = Math.max(maxi, m.get(i));
mini = Math.min(mini, m.get(i));
}
// Printing all the nodes with maximum degree
System.out.print("Nodes with maximum degree : ");
for (int i = 1; i <= n; i++)
{
if (m.get(i) == maxi)
System.out.print(i + " ");
}
System.out.println();
// Printing all the nodes with minimum degree
System.out.print("Nodes with minimum degree : ");
for (int i = 1; i <= n; i++)
{
if (m.get(i) == mini)
System.out.print(i + " ");
}
}
// Driver code
public static void main(String[] args)
{
// Count of nodes and edges
int n = 4, m = 6;
// The edge list
int edges[][] = {{ 1, 2 }, { 1, 3 },
{ 1, 4 }, { 2, 3 },
{ 2, 4 }, { 3, 4 }};
minMax(edges, m, 4);
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 implementation of the approach
# Function to print the nodes having
# maximum and minimum degree
def minMax(edges, leng, n) :
# Map to store the degrees of every node
m = {};
for i in range(leng) :
m[edges[i][0]] = 0;
m[edges[i][1]] = 0;
for i in range(leng) :
# Storing the degree for each node
m[edges[i][0]] += 1;
m[edges[i][1]] += 1;
# maxi and mini variables to store
# the maximum and minimum degree
maxi = 0;
mini = n;
for i in range(1, n + 1) :
maxi = max(maxi, m[i]);
mini = min(mini, m[i]);
# Printing all the nodes
# with maximum degree
print("Nodes with maximum degree : ",
end = "")
for i in range(1, n + 1) :
if (m[i] == maxi) :
print(i, end = " ");
print()
# Printing all the nodes
# with minimum degree
print("Nodes with minimum degree : ",
end = "")
for i in range(1, n + 1) :
if (m[i] == mini) :
print(i, end = " ");
# Driver code
if __name__ == "__main__" :
# Count of nodes and edges
n = 4; m = 6;
# The edge list
edges = [[ 1, 2 ], [ 1, 3 ],
[ 1, 4 ], [ 2, 3 ],
[ 2, 4 ], [ 3, 4 ]];
minMax(edges, m, 4);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to print the nodes having
// maximum and minimum degree
static void minMax(int [,]edges, int len, int n)
{
// Map to store the degrees of every node
Dictionary m = new Dictionary();
for (int i = 0; i < len; i++)
{
// Storing the degree for each node
if(m.ContainsKey(edges[i, 0]))
{
m[edges[i, 0]] = m[edges[i, 0]] + 1;
}
else
{
m.Add(edges[i, 0], 1);
}
if(m.ContainsKey(edges[i, 1]))
{
m[edges[i, 1]] = m[edges[i, 1]] + 1;
}
else
{
m.Add(edges[i, 1], 1);
}
}
// maxi and mini variables to store
// the maximum and minimum degree
int maxi = 0;
int mini = n;
for (int i = 1; i <= n; i++)
{
maxi = Math.Max(maxi, m[i]);
mini = Math.Min(mini, m[i]);
}
// Printing all the nodes with maximum degree
Console.Write("Nodes with maximum degree : ");
for (int i = 1; i <= n; i++)
{
if (m[i] == maxi)
Console.Write(i + " ");
}
Console.WriteLine();
// Printing all the nodes with minimum degree
Console.Write("Nodes with minimum degree : ");
for (int i = 1; i <= n; i++)
{
if (m[i] == mini)
Console.Write(i + " ");
}
}
// Driver code
public static void Main(String[] args)
{
// Count of nodes and edges
int m = 6;
// The edge list
int [,]edges = {{ 1, 2 }, { 1, 3 },
{ 1, 4 }, { 2, 3 },
{ 2, 4 }, { 3, 4 }};
minMax(edges, m, 4);
}
}
// This code is contributed by 29AjayKumar 输出：
Nodes with maximum degree : 1 2 3 4
Nodes with minimum degree : 1 2 3 4