数组中的双毕达哥拉斯三元组

给定一个整数数组arr[] ，任务是检查数组中是否存在孪生勾股三元组。

A Twin Pythagorean Triplet is a Pythagorean triplet (a, b, c) for which two values are consecutive integers. The first few twin triples are (3, 4, 5), (5, 12, 13), (7, 24, 25), (20, 21, 29), (9, 40, 41), (11, 60, 61), (13, 84, 85), (15, 112, 113), …. 

例子：

Input: arr[] = {3, 1, 4, 6, 5} 
Output: True
Explanation: 
There is a Twin Pythagorean triplet (3, 4, 5)

Input: arr[] = {6, 8, 10} 
Output: False 
Explanation:
6² + 8² = 10² 
but in (6, 8, 10) there is no consecutive element. 
So There is no Twin Pythagorean triplet. 

朴素的方法：一个简单的解决方案是运行三个循环，三个循环选择三个数组元素并检查当前的三个元素是否形成孪生勾股三元组。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if there exist
// a twin pythagorean triplet in
// the given array
bool isTriplet(int ar[], int n)
{
    // Loop to check if there is a
    // Pythagorean triplet in the array
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            for (int k = j + 1; k < n; k++) {
 
                // Check if there is
                // consecutive triple
                if (abs(ar[i] - ar[j]) == 1
                    || abs(ar[j] - ar[k]) == 1
                    || abs(ar[i] - ar[k]) == 1) {
 
                    // Calculate square of
                    // array elements
                    int x = ar[i] * ar[i],
                        y = ar[j] * ar[j],
                        z = ar[k] * ar[k];
 
                    if (x == y + z
                        || y == x + z
                        || z == x + y)
                        return true;
                }
            }
        }
    }
 
    return false;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 1, 4, 6, 5 };
    int ar_size = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    isTriplet(arr, ar_size)
        ? cout << "Yes"
        : cout << "No";
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if there exist
// a twin pythagorean triplet in
// the given array
static boolean isTriplet(int ar[], int n)
{
     
    // Loop to check if there is a
    // Pythagorean triplet in the array
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {
            for(int k = j + 1; k < n; k++)
            {
                 
                // Check if there is
                // consecutive triple
                if (Math.abs(ar[i] - ar[j]) == 1 ||
                    Math.abs(ar[j] - ar[k]) == 1 ||
                    Math.abs(ar[i] - ar[k]) == 1)
                {
 
                    // Calculate square of
                    // array elements
                    int x = ar[i] * ar[i],
                        y = ar[j] * ar[j],
                        z = ar[k] * ar[k];
 
                    if (x == y + z ||
                        y == x + z ||
                        z == x + y)
                        return true;
                }
            }
        }
    }
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 3, 1, 4, 6, 5 };
    int ar_size = arr.length;
 
    // Function call
    if(isTriplet(arr, ar_size))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach
 
# Function to check if there exist
# a twin pythagorean triplet in
# the given array
def isTriplet(ar, n):
     
    # Loop to check if there is a
    # Pythagorean triplet in the array
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                 
                # Check if there is
                # consecutive triple
                if (abs(ar[i] - ar[j]) == 1 or
                    abs(ar[j] - ar[k]) == 1 or
                    abs(ar[i] - ar[k]) == 1):
                     
                        # Calculate square of
                        # array elements
                        x = ar[i] * ar[i]
                        y = ar[j] * ar[j]
                        z = ar[k] * ar[k]
                         
                        if (x == y + z or
                            y == x + z or
                            z == x + y):
                            return True
                             
    return False
 
# Driver Code
if __name__=='__main__':
     
    # Given array arr[]
    arr = [ 3, 1, 4, 6, 5 ]
    ar_size = len(arr)
     
    # Function call
    if isTriplet(arr, ar_size):
        print('Yes')
    else:
        print('No')
 
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to check if there exist
// a twin pythagorean triplet in
// the given array
static bool isTriplet(int []ar, int n)
{
     
    // Loop to check if there is a
    // Pythagorean triplet in the array
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {
            for(int k = j + 1; k < n; k++)
            {
                 
                // Check if there is
                // consecutive triple
                if (Math.Abs(ar[i] - ar[j]) == 1 ||
                    Math.Abs(ar[j] - ar[k]) == 1 ||
                    Math.Abs(ar[i] - ar[k]) == 1)
                {
 
                    // Calculate square of
                    // array elements
                    int x = ar[i] * ar[i],
                        y = ar[j] * ar[j],
                        z = ar[k] * ar[k];
 
                    if (x == y + z ||
                        y == x + z ||
                        z == x + y)
                        return true;
                }
            }
        }
    }
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 3, 1, 4, 6, 5 };
    int ar_size = arr.Length;
 
    // Function call
    if(isTriplet(arr, ar_size))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// Javascript program for the above approach
 
// Function to check if there exist
// a twin pythagorean triplet in
// the given array
function isTriplet(ar, n)
{
    // Loop to check if there is a
    // Pythagorean triplet in the array
    for (var i = 0; i < n; i++) {
        for (var j = i + 1; j < n; j++) {
 
            for (var k = j + 1; k < n; k++) {
 
                // Check if there is
                // consecutive triple
                if (Math.abs(ar[i] - ar[j]) == 1
                    || Math.abs(ar[j] - ar[k]) == 1
                    || Math.abs(ar[i] - ar[k]) == 1) {
 
                    // Calculate square of
                    // array elements
                    var x = ar[i] * ar[i],
                        y = ar[j] * ar[j],
                        z = ar[k] * ar[k];
 
                    if (x == y + z
                        || y == x + z
                        || z == x + y)
                        return true;
                }
            }
        }
    }
 
    return false;
}
 
// Driver Code
 
// Given array arr[]
var arr = [ 3, 1, 4, 6, 5 ];
var ar_size = arr.length;
 
// Function Call
isTriplet(arr, ar_size)
    ? document.write( "Yes")
    : document.write( "No");
 
// This code is contributed by itsok.

输出：

Yes

时间复杂度： O(N ³ )
辅助村： O(1)

有效的方法：

• 对输入数组中的每个元素进行平方。这一步需要 O(n) 时间。
• 按升序对平方数组进行排序。这一步需要 O(nLogn) 时间。
• 要找到一个三元组 (a, b, c) 使得a ² = b ² + c ²并且至少一对是连续的，请执行以下操作。
  • 将 ‘a’ 固定为排序数组的最后一个元素。
  • 现在在第一个元素和“a”之间的子数组中搜索对 (b, c)。可以使用本文方法 1 中讨论的中间算法在 O(n) 时间内找到具有给定总和的对 (b, c)。
  • 如果没有找到当前 ‘a’ 的配对，则将 ‘a’ 向后移动一个位置并重复上一步。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
 
using namespace std;
 
// Function to check if any of the
// two numbers are consecutive
// in the given triplet
bool isConsective(int a, int b, int c)
{
    return abs(a - b) == 1
           || abs(b - c) == 1
           || abs(c - a) == 1;
}
 
// Function to check if there exist a
// pythagorean triplet in the array
bool isTriplet(int arr[], int n)
{
 
    // Square array elements
    for (int i = 0; i < n; i++)
        arr[i] = arr[i] * arr[i];
 
    // Sort array elements
    sort(arr, arr + n);
 
    // Now fix one element one by one
    // and find the other two elements
    for (int i = n - 1; i >= 2; i--) {
 
        // To find the other two elements,
        // start two index variables from
        // two corners of the array and move
        // them toward each other
        int l = 0;
        int r = i - 1;
 
        while (l < r) {
 
            // A triplet found
            if (arr[l] + arr[r] == arr[i]
                && isConsective(sqrt(arr[l]),
                                sqrt(arr[r]),
                                sqrt(arr[i]))) {
 
                return true;
            }
 
            // Else either move 'l' or 'r'
            (arr[l] + arr[r] < arr[i])
                ? l++
                : r--;
        }
    }
 
    // If we reach here,
    // then no triplet found
    return false;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 1, 4, 6, 5 };
    int arr_size = sizeof(arr)
                   / sizeof(arr[0]);
 
    // Function Call
    isTriplet(arr, arr_size)
        ? cout << "Yes"
        : cout << "No";
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if any of the
// two numbers are consecutive
// in the given triplet
static boolean isConsective(int a, int b, int c)
{
    return Math.abs(a - b) == 1 ||
           Math.abs(b - c) == 1 ||
           Math.abs(c - a) == 1;
}
 
// Function to check if there exist a
// pythagorean triplet in the array
static boolean isTriplet(int arr[], int n)
{
 
    // Square array elements
    for(int i = 0; i < n; i++)
        arr[i] = arr[i] * arr[i];
 
    // Sort array elements
    Arrays.sort(arr);
 
    // Now fix one element one by one
    // and find the other two elements
    for(int i = n - 1; i >= 2; i--)
    {
 
        // To find the other two elements,
        // start two index variables from
        // two corners of the array and move
        // them toward each other
        int l = 0;
        int r = i - 1;
 
        while (l < r)
        {
             
            // A triplet found
            if (arr[l] + arr[r] == arr[i] &&
                isConsective((int)Math.sqrt(arr[l]),
                             (int)Math.sqrt(arr[r]),
                             (int)Math.sqrt(arr[i])))
            {
                return true;
            }
 
            // Else either move 'l' or 'r'
            if(arr[l] + arr[r] < arr[i])
                l++;
            else
                r--;
        }
    }
 
    // If we reach here,
    // then no triplet found
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 3, 1, 4, 6, 5 };
    int arr_size = arr.length;
 
    // Function call
    if(isTriplet(arr, arr_size))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by gauravrajput1

蟒蛇3

# Python3 program for the above approach
import math
 
# Function to check if any of the
# two numbers are consecutive
# in the given triplet
def isConsective(a, b, c):
     
    return bool(abs(a - b) == 1 or
                abs(b - c) == 1 or
                abs(c - a) == 1)
   
# Function to check if there exist a
# pythagorean triplet in the array
def isTriplet(arr, n):
   
    # Square array elements
    for i in range(n):
        arr[i] = arr[i] * arr[i]
   
    # Sort array elements
    arr.sort()
   
    # Now fix one element one by one
    # and find the other two elements
    for i in range(n - 1, 1, -1):
   
        # To find the other two elements,
        # start two index variables from
        # two corners of the array and move
        # them toward each other
        l = 0
        r = i - 1
   
        while (l < r):
   
            # A triplet found
            if (arr[l] + arr[r] == arr[i] and
                isConsective(math.sqrt(arr[l]),
                             math.sqrt(arr[r]),
                             math.sqrt(arr[i]))):
   
                return bool(True)
   
            # Else either move 'l' or 'r'
            if(arr[l] + arr[r] < arr[i]):
                l = l + 1
            else:
                r = r - 1
   
    # If we reach here,
    # then no triplet found
    return bool(False)
     
# Driver code
 
# Given array arr[]
arr = [ 3, 1, 4, 6, 5 ]
arr_size = len(arr)
 
# Function call
if(isTriplet(arr, arr_size)):
    print("Yes")
else:
    print("No")
 
# This code is contributed divyeshrabadiya07

C＃

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if any of the
// two numbers are consecutive
// in the given triplet
static bool isConsective(int a, int b, int c)
{
    return Math.Abs(a - b) == 1 ||
           Math.Abs(b - c) == 1 ||
           Math.Abs(c - a) == 1;
}
 
// Function to check if there exist a
// pythagorean triplet in the array
static bool isTriplet(int []arr, int n)
{
 
    // Square array elements
    for(int i = 0; i < n; i++)
        arr[i] = arr[i] * arr[i];
 
    // Sort array elements
    Array.Sort(arr);
 
    // Now fix one element one by one
    // and find the other two elements
    for(int i = n - 1; i >= 2; i--)
    {
         
        // To find the other two elements,
        // start two index variables from
        // two corners of the array and move
        // them toward each other
        int l = 0;
        int r = i - 1;
 
        while (l < r)
        {
             
            // A triplet found
            if (arr[l] + arr[r] == arr[i] &&
                isConsective((int)Math.Sqrt(arr[l]),
                             (int)Math.Sqrt(arr[r]),
                             (int)Math.Sqrt(arr[i])))
            {
                return true;
            }
 
            // Else either move 'l' or 'r'
            if(arr[l] + arr[r] < arr[i])
                l++;
            else
                r--;
        }
    }
 
    // If we reach here,
    // then no triplet found
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 3, 1, 4, 6, 5 };
    int arr_size = arr.Length;
 
    // Function call
    if(isTriplet(arr, arr_size))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by gauravrajput1

Javascript

// JavaScript program for the above approach
 
// Function to check if any of the
// two numbers are consecutive
// in the given triplet
function isConsective( a, b, c){
    return Math.abs(a - b) == 1
           || Math.abs(b - c) == 1
           || Math.abs(c - a) == 1;
}
 
// Function to check if there exist a
// pythagorean triplet in the array
function isTriplet( arr, n){
    // Square array elements
    for (let i = 0; i < n; i++)
        arr[i] = arr[i] * arr[i];
 
    // Sort array elements
    arr.sort(function(a, b){return a - b});
 
    // Now fix one element one by one
    // and find the other two elements
    for (let i = n - 1; i >= 2; i--) {
 
        // To find the other two elements,
        // start two index variables from
        // two corners of the array and move
        // them toward each other
        let l = 0;
        let r = i - 1;
 
        while (l < r) {
 
            // A triplet found
            if (arr[l] + arr[r] == arr[i]
                && isConsective(Math.sqrt(arr[l]),
                                Math.sqrt(arr[r]),
                                Math.sqrt(arr[i]))) {
 
                return true;
            }
 
            // Else either move 'l' or 'r'
            (arr[l] + arr[r] < arr[i])
                ? l++
                : r--;
        }
    }
 
    // If we reach here,
    // then no triplet found
    return false;
}
 
// Driver Code
// Given array arr[]
let arr = [ 3, 1, 4, 6, 5 ];
let arr_size = arr.length;
 
// Function Call
isTriplet(arr, arr_size)
        ?document.write( "Yes")
        : document.write( "No");

输出：

Yes

时间复杂度： O(N ² )
辅助空间： O(1)

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