Java程序打印形成 AP 的有序数组中的所有三元组
给定一个不同正整数的排序数组，打印所有形成 AP（或算术级数）的三元组
例子 ：
Input : arr[] = { 2, 6, 9, 12, 17, 22, 31, 32, 35, 42 };
Output :
6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42
Input : arr[] = { 3, 5, 6, 7, 8, 10, 12};
Output :
3 5 7
5 6 7
6 7 8
6 8 10
8 10 12一个简单的解决方案是运行三个嵌套循环来生成所有三元组，并为每个三元组检查它是否形成 AP。该解决方案的时间复杂度为 O(n 3 )
更好的解决方案是使用散列。我们从左到右遍历数组。我们将每个元素视为中间元素，将其之后的所有元素视为下一个元素。为了搜索前一个元素，我们使用哈希表。
Java
// Java program to print all
// triplets in given array
// that form Arithmetic
// Progression
import java.io.*;
import java.util.*;
class GFG
{
// Function to print
// all triplets in
// given sorted array
// that forms AP
static void printAllAPTriplets(int []arr,
int n)
{
ArrayList s =
new ArrayList();
for (int i = 0;
i < n - 1; i++)
{
for (int j = i + 1; j < n; j++)
{
// Use hash to find if
// there is a previous
// element with difference
// equal to arr[j] - arr[i]
int diff = arr[j] - arr[i];
boolean exists =
s.contains(arr[i] - diff);
if (exists)
System.out.println(arr[i] - diff +
" " + arr[i] +
" " + arr[j]);
}
s.add(arr[i]);
}
}
// Driver code
public static void main(String args[])
{
int []arr = {2, 6, 9, 12, 17,
22, 31, 32, 35, 42};
int n = arr.length;
printAllAPTriplets(arr, n);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1) Java
// Java implementation to print
// all the triplets in given array
// that form Arithmetic Progression
import java.io.*;
class GFG
{
// Function to print all triplets in
// given sorted array that forms AP
static void findAllTriplets(int arr[], int n)
{
for (int i = 1; i < n - 1; i++)
{
// Search other two elements
// of AP with arr[i] as middle.
for (int j = i - 1, k = i + 1; j >= 0 && k < n;)
{
// if a triplet is found
if (arr[j] + arr[k] == 2 * arr[i])
{
System.out.println(arr[j] +" " +
arr[i]+ " " + arr[k]);
// Since elements are distinct,
// arr[k] and arr[j] cannot form
// any more triplets with arr[i]
k++;
j--;
}
// If middle element is more move to
// higher side, else move lower side.
else if (arr[j] + arr[k] < 2 * arr[i])
k++;
else
j--;
}
}
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 };
int n = arr.length;
findAllTriplets(arr, n);
}
}
// This code is contributed by vt_m.输出 ：
6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42时间复杂度： O(n 2 )
辅助空间： O(n)
一个有效的解决方案是基于数组已排序的事实。我们使用与 GP 三元组问题中讨论的相同概念。这个想法是从第二个元素开始，将每个元素固定为中间元素，并在三元组中搜索其他两个元素（一个小一个大）。
下面是上述思想的实现。
Java
// Java implementation to print
// all the triplets in given array
// that form Arithmetic Progression
import java.io.*;
class GFG
{
// Function to print all triplets in
// given sorted array that forms AP
static void findAllTriplets(int arr[], int n)
{
for (int i = 1; i < n - 1; i++)
{
// Search other two elements
// of AP with arr[i] as middle.
for (int j = i - 1, k = i + 1; j >= 0 && k < n;)
{
// if a triplet is found
if (arr[j] + arr[k] == 2 * arr[i])
{
System.out.println(arr[j] +" " +
arr[i]+ " " + arr[k]);
// Since elements are distinct,
// arr[k] and arr[j] cannot form
// any more triplets with arr[i]
k++;
j--;
}
// If middle element is more move to
// higher side, else move lower side.
else if (arr[j] + arr[k] < 2 * arr[i])
k++;
else
j--;
}
}
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 };
int n = arr.length;
findAllTriplets(arr, n);
}
}
// This code is contributed by vt_m.
输出 ：
6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42时间复杂度： O(n 2 )
辅助空间： O(1)
有关详细信息，请参阅完整的文章在形成 AP 的排序数组中打印所有三元组！