给定一个表示二叉树的节点的二维数组arr [] [] ，任务是在不实际构造此树的情况下，找到该树的兄弟姐妹的最大GCD。
例子：
Input: arr[][] = {{4, 5}, {4, 2}, {2, 3}, {2, 1}, {3, 6}, {3, 12}}
Output: 6
Explanation:
For the above tree, the maximum GCD for the sibilings is formed for the nodes 6 and 12 for the children of node 3.
Input: arr[][] = {{5, 4}, {5, 8}, {4, 6}, {4, 9}, {8, 10}, {10, 20}, {10, 30}}
Output: 10
方法：想法是形成一个向量，并以该向量的形式存储树。将树以矢量形式存储后，会发生以下情况：
- 如果向量大小为0或1 ，则打印0，因为找不到GCD。
- 对于所有其他情况，由于我们以成对的形式存储树，因此我们考虑了两个对的第一个值并进行比较。但是首先，您需要对这对边进行排序。
例如，假设向量A和B中有两对。我们检查是否：
A.first == B.first- 如果它们都匹配，则它们都属于同一个父对象。因此，我们计算该对中第二个值的GCD，最后打印出所有此类GCD的最大值。
下面是上述方法的实现：
C++
// C++ program to find the maximum
// GCD of the siblings of a binary tree
#include
using namespace std;
// Function to find maximum GCD
int max_gcd(vector >& v)
{
// No child or Single child
if (v.size() == 1 || v.size() == 0)
return 0;
sort(v.begin(), v.end());
// To get the first pair
pair a = v[0];
pair b;
int ans = INT_MIN;
for (int i = 1; i < v.size(); i++) {
b = v[i];
// If both the pairs belongs to
// the same parent
if (b.first == a.first)
// Update ans with the max
// of current gcd and
// gcd of both children
ans
= max(ans,
__gcd(a.second,
b.second));
// Update previous
// for next iteration
a = b;
}
return ans;
}
// Driver function
int main()
{
vector > v;
v.push_back(make_pair(5, 4));
v.push_back(make_pair(5, 8));
v.push_back(make_pair(4, 6));
v.push_back(make_pair(4, 9));
v.push_back(make_pair(8, 10));
v.push_back(make_pair(10, 20));
v.push_back(make_pair(10, 30));
cout << max_gcd(v);
return 0;
} Java
// Java program to find the maximum
// GCD of the siblings of a binary tree
import java.util.*;
import java.lang.*;
class GFG{
// Function to find maximum GCD
static int max_gcd(ArrayList v)
{
// No child or Single child
if (v.size() == 1 || v.size() == 0)
return 0;
Collections.sort(v, new Comparator() {
public int compare(int[] a, int[] b) {
return a[0]-b[0];
}
});
// To get the first pair
int[] a = v.get(0);
int[] b = new int[2];
int ans = Integer.MIN_VALUE;
for(int i = 1; i < v.size(); i++)
{
b = v.get(i);
// If both the pairs belongs to
// the same parent
if (b[0] == a[0])
// Update ans with the max
// of current gcd and
// gcd of both children
ans = Math.max(ans,
gcd(a[1],
b[1]));
// Update previous
// for next iteration
a = b;
}
return ans;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver code
public static void main(String[] args)
{
ArrayList v = new ArrayList<>();
v.add(new int[]{5, 4});
v.add(new int[]{5, 8});
v.add(new int[]{4, 6});
v.add(new int[]{4, 9});
v.add(new int[]{8, 10});
v.add(new int[]{10, 20});
v.add(new int[]{10, 30});
System.out.println(max_gcd(v));
}
}
// This code is contributed by offbeat Python3
# Python3 program to find the maximum
# GCD of the siblings of a binary tree
from math import gcd
# Function to find maximum GCD
def max_gcd(v):
# No child or Single child
if (len(v) == 1 or len(v) == 0):
return 0
v.sort()
# To get the first pair
a = v[0]
ans = -10**9
for i in range(1, len(v)):
b = v[i]
# If both the pairs belongs to
# the same parent
if (b[0] == a[0]):
# Update ans with the max
# of current gcd and
# gcd of both children
ans = max(ans, gcd(a[1], b[1]))
# Update previous
# for next iteration
a = b
return ans
# Driver function
if __name__ == '__main__':
v=[]
v.append([5, 4])
v.append([5, 8])
v.append([4, 6])
v.append([4, 9])
v.append([8, 10])
v.append([10, 20])
v.append([10, 30])
print(max_gcd(v))
# This code is contributed by mohit kumar 29C#
// C# program to find the maximum
// GCD of the siblings of a binary tree
using System.Collections;
using System;
class GFG{
// Function to find maximum GCD
static int max_gcd(ArrayList v)
{
// No child or Single child
if (v.Count == 1 || v.Count == 0)
return 0;
v.Sort();
// To get the first pair
int[] a = (int [])v[0];
int[] b = new int[2];
int ans = -10000000;
for(int i = 1; i < v.Count; i++)
{
b = (int[])v[i];
// If both the pairs belongs to
// the same parent
if (b[0] == a[0])
// Update ans with the max
// of current gcd and
// gcd of both children
ans = Math.Max(ans, gcd(a[1], b[1]));
// Update previous
// for next iteration
a = b;
}
return ans;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver code
public static void Main(string[] args)
{
ArrayList v = new ArrayList();
v.Add(new int[]{5, 4});
v.Add(new int[]{5, 8});
v.Add(new int[]{4, 6});
v.Add(new int[]{4, 9});
v.Add(new int[]{8, 10});
v.Add(new int[]{10, 20});
v.Add(new int[]{10, 30});
Console.Write(max_gcd(v));
}
}
// This code is contributed by rutvik_56输出：
10