给定具有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，节点的度数是入射到其上的边的数量，因此可以通过在边列表中计算其频率来计算每个节点的度。因此，该方法是使用地图从边缘列表计算每个顶点的频率，并使用地图查找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
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