数组中的毕达哥拉斯三元组

给定一个整数数组，编写一个函数，如果存在满足 a ² + b ² = c ²的三元组 (a, b, c)，则返回 true。

示例：

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

Input: arr[] = {10, 4, 6, 12, 5} 
Output: False 
There is no Pythagorean triplet. 

方法1（天真）
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，并检查当前三个元素是否形成一个毕达哥拉斯三元组。

下面是上述想法的实现：

C++

// A C++ program that returns true if there is a Pythagorean
// Triplet in a given array.
#include 
 
using namespace std;
 
// Returns true if there is Pythagorean triplet in ar[0..n-1]
bool isTriplet(int ar[], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = j + 1; k < n; k++) {
                // 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;
            }
        }
    }
 
    // If we reach here, no triplet found
    return false;
}
 
/* Driver program to test above function */
int main()
{
    int ar[] = { 3, 1, 4, 6, 5 };
    int ar_size = sizeof(ar) / sizeof(ar[0]);
    isTriplet(ar, ar_size) ? cout << "Yes" : cout << "No";
    return 0;
}

Java

// A Java program that returns true if there is a Pythagorean
// Triplet in a given array.
import java.io.*;
 
class PythagoreanTriplet {
 
    // Returns true if there is Pythagorean triplet in ar[0..n-1]
    static boolean isTriplet(int ar[], int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {
                    // 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;
                }
            }
        }
 
        // If we reach here, no triplet found
        return false;
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        int ar[] = { 3, 1, 4, 6, 5 };
        int ar_size = ar.length;
        if (isTriplet(ar, ar_size) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
/* This code is contributed by Devesh Agrawal */

Python3

# Python program to check if there is Pythagorean
# triplet in given array
 
# Returns true if there is Pythagorean
# triplet in ar[0..n-1]
 
def isTriplet(ar, n):
    j = 0
     
    for i in range(n - 2):
        for k in range(j + 1, n):
            for j in range(i + 1, n - 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 1
     
    # If we reach here, no triplet found
    return 0
 
 
# Driver program to test above function
ar = [3, 1, 4, 6, 5]
ar_size = len(ar)
 
if(isTriplet(ar, ar_size)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Aditi Sharma

C#

// A C# program that returns true
// if there is a Pythagorean
// Triplet in a given array.
using System;
 
class GFG {
 
    // Returns true if there is Pythagorean
    // triplet in ar[0..n-1]
    static bool isTriplet(int[] ar, int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {
 
                    // 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;
                }
            }
        }
 
        // If we reach here,
        // no triplet found
        return false;
    }
 
    // Driver code
    public static void Main()
    {
        int[] ar = { 3, 1, 4, 6, 5 };
        int ar_size = ar.Length;
        if (isTriplet(ar, ar_size) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by shiv_bhakt.

Javascript

 
// A Javascript program that returns
// true if there is a Pythagorean
// Triplet in a given array.
 
// Returns true if there is
// Pythagorean triplet in ar[0..n-1]
function isTriplet( ar, n)
{
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            for (let k = j + 1; k < n; k++) {
          // Calculate square of array elements
                let 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;
            }
        }
    }
 
    // If we reach here, no triplet found
    return false;
}
 
 
    // driver code
     
    let ar = [ 3, 1, 4, 6, 5 ];
    let ar_size = ar.length;
    if (isTriplet(ar, ar_size) == true)
        document.write("Yes");
    else
        document.write("No");
     

输出：

Yes

上述解决方案的时间复杂度为 O(n ³ )。

方法2（使用排序）
我们可以通过首先对数组进行排序在 O(n ² ) 时间内解决这个问题。
1) 对输入数组中的每个元素进行平方。这一步需要 O(n) 时间。
2) 按升序对方阵进行排序。这一步需要 O(nLogn) 时间。
3) 要找到满足 a ² = b ² + c ²的三元组 (a, b, c)，请执行以下操作。

1. 将“a”固定为排序数组的最后一个元素。
2. 现在在第一个元素和“a”之间的子数组中搜索对 (b, c)。使用本文方法 1 中讨论的中间相遇算法，可以在 O(n) 时间内找到具有给定总和的对 (b, c)。
3. 如果没有找到当前“a”的对，则将“a”向后移动一个位置并重复步骤 3.2。

下图是上述方法的试运行：

下面是上述方法的实现：

C++

// A C++ program that returns true if there is a Pythagorean
// Triplet in a given array.
#include 
#include 
 
using namespace std;
 
// Returns true if there is a triplet with following property
// A[i]*A[i] = A[j]*A[j] + A[k]*[k]
// Note that this function modifies given 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; // index of the first element in arr[0..i-1]
        int r = i - 1; // index of the last element in arr[0..i-1]
        while (l < r) {
            // A triplet found
            if (arr[l] + arr[r] == 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 program to test above function */
int main()
{
    int arr[] = { 3, 1, 4, 6, 5 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    isTriplet(arr, arr_size) ? cout << "Yes" : cout << "No";
    return 0;
}

Java

// A Java program that returns true if there is a Pythagorean
// Triplet in a given array.
import java.io.*;
import java.util.*;
 
class PythagoreanTriplet {
    // Returns true if there is a triplet with following property
    // A[i]*A[i] = A[j]*A[j] + A[k]*[k]
    // Note that this function modifies given 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; // index of the first element in arr[0..i-1]
            int r = i - 1; // index of the last element in arr[0..i-1]
            while (l < r) {
                // A triplet found
                if (arr[l] + arr[r] == 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 program to test above function
    public static void main(String[] args)
    {
        int arr[] = { 3, 1, 4, 6, 5 };
        int arr_size = arr.length;
        if (isTriplet(arr, arr_size) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
/*This code is contributed by Devesh Agrawal*/

Python3

# Python program that returns true if there is
# a Pythagorean Triplet in a given array.
 
# Returns true if there is Pythagorean
# triplet in ar[0..n-1]
def isTriplet(ar, n):
    # Square all the elements
    for i in range(n):
        ar[i] = ar[i] * ar[i]
 
    # sort array elements
    ar.sort()
 
    # fix one element
    # and find other two
    # i goes from n - 1 to 2
    for i in range(n-1, 1, -1):
        # start two index variables from
        # two corners of the array and
        # move them toward each other
        j = 0
        k = i - 1
        while (j < k):
            # A triplet found
            if (ar[j] + ar[k] == ar[i]):
                return True
            else:
                if (ar[j] + ar[k] < ar[i]):
                    j = j + 1
                else:
                    k = k - 1
    # If we reach here, then no triplet found
    return False
 
# Driver program to test above function */
ar = [3, 1, 4, 6, 5]
ar_size = len(ar)
if(isTriplet(ar, ar_size)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Aditi Sharma

C＃

// C# program that returns true
// if there is a Pythagorean
// Triplet in a given array.
using System;
 
class GFG {
 
    // Returns true if there is a triplet
    // with following property A[i]*A[i]
    // = A[j]*A[j]+ A[k]*[k] Note that
    // this function modifies given 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
            // index of the first element
            // in arr[0..i-1]
            int l = 0;
 
            // index of the last element
            // in arr[0..i - 1]
            int r = i - 1;
            while (l < r) {
 
                // A triplet found
                if (arr[l] + arr[r] == 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()
    {
        int[] arr = { 3, 1, 4, 6, 5 };
        int arr_size = arr.Length;
        if (isTriplet(arr, arr_size) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by shiv_bhakt.

PHP

= 2; $i--)
    {
         
        // To find the other two
        // elements, start two index
        // variables from two corners
        // of the array and move
        // them toward each other
         
        // index of the first element
        // in arr[0..i-1]
        $l = 0;
         
        // index of the last element
        // in arr[0..i-1]
        $r = $i - 1;
        while ($l < $r)
        {
             
            // A triplet found
            if ($arr[$l] + $arr[$r] == $arr[$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
    $arr = array(3, 1, 4, 6, 5);
    $arr_size = count($arr);
    if(isTriplet($arr, $arr_size))
        echo "Yes";
    else
        echo "No";
 
// This code is contributed by anuj_67.
?>

Javascript

// A javascript program that returns true if there is a Pythagorean
// Triplet in a given array.
 
    // Returns true if there is a triplet with following property
    // A[i]*A[i] = A[j]*A[j] + A[k]*[k]
    // Note that this function modifies given array
    function isTriplet(arr , n)
    {
     
        // Square array elements
        for (i = 0; i < n; i++)
            arr[i] = arr[i] * arr[i];
 
        // Sort array elements
        arr.sort((a,b)=>a-b);
 
        // Now fix one element one by one and find the other two
        // elements
        for (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
            var l = 0; // index of the first element in arr[0..i-1]
            var r = i - 1; // index of the last element in arr[0..i-1]
            while (l < r)
            {
             
                // A triplet found
                if (arr[l] + arr[r] == 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 program to test above function   
        var arr = [ 3, 1, 4, 6, 5 ];
        var arr_size = arr.length;
        if (isTriplet(arr, arr_size) == true)
            document.write("Yes");
        else
            document.write("No");
 
// This code is contributed by umadevi9616

输出：

Yes 

该方法的时间复杂度为 O(n ² )。

方法3 ：（使用散列）
这个问题也可以使用散列来解决。我们可以使用哈希映射来标记给定数组的所有值。使用两个循环，我们可以迭代 a 和 b 的所有可能组合，然后检查是否存在第三个值 c。如果存在任何这样的值，则存在毕达哥拉斯三元组。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
// Function to check if the
// Pythagorean triplet exists or not
bool checkTriplet(int arr[], int n)
{
    int maximum = 0;
 
    // Find the maximum element
    for (int i = 0; i < n; i++) {
        maximum = max(maximum, arr[i]);
    }
 
    // Hashing array
    int hash[maximum + 1] = { 0 };
 
    // Increase the count of array elements
    // in hash table
    for (int i = 0; i < n; i++)
        hash[arr[i]]++;
 
    // Iterate for all possible a
    for (int i = 1; i < maximum + 1; i++) {
 
        // If a is not there
        if (hash[i] == 0)
            continue;
 
        // Iterate for all possible b
        for (int j = 1; j < maximum + 1; j++) {
 
            // If a and b are same and there is only one a
            // or if there is no b in original array
            if ((i == j && hash[i] == 1) || hash[j] == 0)
                continue;
 
            // Find c
            int val = sqrt(i * i + j * j);
 
            // If c^2 is not a perfect square
            if ((val * val) != (i * i + j * j))
                continue;
 
            // If c exceeds the maximum value
            if (val > maximum)
                continue;
 
            // If there exists c in the original array,
            // we have the triplet
            if (hash[val]) {
                return true;
            }
        }
    }
    return false;
}
// Driver Code
int main()
{
    int arr[] = { 3, 2, 4, 6, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (checkTriplet(arr, n))
        cout << "Yes";
    else
        cout << "No";
}

Java

import java.util.*;
 
class GFG
{
 
// Function to check if the
// Pythagorean triplet exists or not
static boolean checkTriplet(int arr[], int n)
{
    int maximum = 0;
 
    // Find the maximum element
    for (int i = 0; i < n; i++)
    {
        maximum = Math.max(maximum, arr[i]);
    }
 
    // Hashing array
    int []hash = new int[maximum + 1];
 
    // Increase the count of array elements
    // in hash table
    for (int i = 0; i < n; i++)
        hash[arr[i]]++;
 
    // Iterate for all possible a
    for (int i = 1; i < maximum + 1; i++)
    {
 
        // If a is not there
        if (hash[i] == 0)
            continue;
 
        // Iterate for all possible b
        for (int j = 1; j < maximum + 1; j++)
        {
 
            // If a and b are same and there is only one a
            // or if there is no b in original array
            if ((i == j && hash[i] == 1) || hash[j] == 0)
                continue;
 
            // Find c
            int val = (int) Math.sqrt(i * i + j * j);
 
            // If c^2 is not a perfect square
            if ((val * val) != (i * i + j * j))
                continue;
 
            // If c exceeds the maximum value
            if (val > maximum)
                continue;
 
            // If there exists c in the original array,
            // we have the triplet
            if (hash[val] == 1)
            {
                return true;
            }
        }
    }
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 4, 6, 5 };
    int n = arr.length;
    if (checkTriplet(arr, n))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Function to check if the
# Pythagorean triplet exists or not
import math
 
def checkTriplet(arr, n):
    maximum = 0
 
    # Find the maximum element
    maximum = max(arr)
 
        # Hashing array
    hash = [0]*(maximum+1)
 
    # Increase the count of array elements
    # in hash table
    for i in range(n):
        hash[arr[i]] += 1
 
        # Iterate for all possible a
    for i in range(1, maximum+1):
        # If a is not there
        if (hash[i] == 0):
            continue
 
        # Iterate for all possible b
        for j in range(1, maximum+1):
            # If a and b are same and there is only one a
            # or if there is no b in original array
            if ((i == j and hash[i] == 1) or hash[j] == 0):
                continue
 
            # Find c
            val = int(math.sqrt(i * i + j * j))
 
            # If c^2 is not a perfect square
            if ((val * val) != (i * i + j * j)):
                continue
 
            # If c exceeds the maximum value
            if (val > maximum):
                continue
 
            # If there exists c in the original array,
            # we have the triplet
            if (hash[val]):
                return True
    return False
 
 
# Driver Code
arr = [3, 2, 4, 6, 5]
n = len(arr)
if (checkTriplet(arr, n)):
    print("Yes")
else:
    print("No")
 
 
# This code is contributed by ankush_953

C＃

using System;
 
class GFG
{
 
// Function to check if the
// Pythagorean triplet exists or not
static bool checkTriplet(int []arr, int n)
{
    int maximum = 0;
 
    // Find the maximum element
    for (int i = 0; i < n; i++)
    {
        maximum = Math.Max(maximum, arr[i]);
    }
 
    // Hashing array
    int []hash = new int[maximum + 1];
 
    // Increase the count of array elements
    // in hash table
    for (int i = 0; i < n; i++)
        hash[arr[i]]++;
 
    // Iterate for all possible a
    for (int i = 1; i < maximum + 1; i++)
    {
 
        // If a is not there
        if (hash[i] == 0)
            continue;
 
        // Iterate for all possible b
        for (int j = 1; j < maximum + 1; j++)
        {
 
            // If a and b are same and there is only one a
            // or if there is no b in original array
            if ((i == j && hash[i] == 1) || hash[j] == 0)
                continue;
 
            // Find c
            int val = (int) Math.Sqrt(i * i + j * j);
 
            // If c^2 is not a perfect square
            if ((val * val) != (i * i + j * j))
                continue;
 
            // If c exceeds the maximum value
            if (val > maximum)
                continue;
 
            // If there exists c in the original array,
            // we have the triplet
            if (hash[val] == 1)
            {
                return true;
            }
        }
    }
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 4, 6, 5 };
    int n = arr.Length;
    if (checkTriplet(arr, n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
    // Function to check if the
    // Pythagorean triplet exists or not
    function checkTriplet(arr , n) {
        var maximum = 0;
 
        // Find the maximum element
        for (i = 0; i < n; i++) {
            maximum = Math.max(maximum, arr[i]);
        }
 
        // Hashing array
        var hash = Array(maximum + 1).fill(0);
 
        // Increase the count of array elements
        // in hash table
        for (i = 0; i < n; i++)
            hash[arr[i]]++;
 
        // Iterate for all possible a
        for (i = 1; i < maximum + 1; i++) {
 
            // If a is not there
            if (hash[i] == 0)
                continue;
 
            // Iterate for all possible b
            for (j = 1; j < maximum + 1; j++) {
 
                // If a and b are same and there is only one a
                // or if there is no b in original array
                if ((i == j && hash[i] == 1) || hash[j] == 0)
                    continue;
 
                // Find c
                var val = parseInt( Math.sqrt(i * i + j * j));
 
                // If c^2 is not a perfect square
                if ((val * val) != (i * i + j * j))
                    continue;
 
                // If c exceeds the maximum value
                if (val > maximum)
                    continue;
 
                // If there exists c in the original array,
                // we have the triplet
                if (hash[val] == 1) {
                    return true;
                }
            }
        }
        return false;
    }
 
    // Driver Code
        var arr = [ 3, 2, 4, 6, 5 ];
        var n = arr.length;
        if (checkTriplet(arr, n))
            document.write("Yes");
        else
            document.write("No");
 
// This code is contributed by gauravrajput1
 

输出

Yes

感谢奋斗者提出上述方法。
时间复杂度：O( max * max )，其中 max 是数组中最大的元素。

方法-4：使用STL

方法：

该问题可以使用有序映射和无序映射来解决。无需以有序方式存储元素，因此无序映射的实现速度更快。我们可以使用无序映射来标记给定数组的所有值。使用两个循环，我们可以迭代 a 和 b 的所有可能组合，然后检查是否存在第三个值 c。如果存在任何这样的值，则存在毕达哥拉斯三元组。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
//  Returns true if there is Pythagorean triplet in
//  ar[0..n-1]
bool checkTriplet(int arr[], int n)
{
    // initializing unordered map with key and value as
    // integers
    unordered_map umap;
     
    // Increase the count of array elements in unordered map
    for (int i = 0; i < n; i++)
        umap[arr[i]] = umap[arr[i]] + 1;
 
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {   
            // calculating the squares of two elements as
            // integer and float
            int p = sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
            float q
                = sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
             
            // Condition is true if the value is same in
            // integer and float and also the value is
            // present in unordered map
            if (p == q && umap[p] != 0)
                return true;
        }
    }
    
    // If we reach here, no triplet found
    return false;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 4, 6, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (checkTriplet(arr, n))
        cout << "Yes";
    else
        cout << "No";
}
// This code is contributed by Vikkycirus

Java

import java.util.*;
class GFG{
 
  //  Returns true if there is Pythagorean triplet in
  //  ar[0..n-1]
  static boolean checkTriplet(int arr[], int n)
  {
 
    // initializing unordered map with key and value as
    // integers
    HashMap umap = new HashMap<>();
 
    // Increase the count of array elements in unordered map
    for (int i = 0; i < n; i++)
      if(umap.containsKey(arr[i]))
        umap.put(arr[i] , umap.get(arr[i]) + 1);
    else
      umap.put(arr[i], 1);
 
    for (int i = 0; i < n - 1; i++)
    {
      for (int j = i + 1; j < n; j++)
      {   
 
        // calculating the squares of two elements as
        // integer and float
        int p =(int) Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
        float q
          =(float) Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
 
        // Condition is true if the value is same in
        // integer and float and also the value is
        // present in unordered map
        if (p == q && umap.get(p) != 0)
          return true;
      }
    }
 
    // If we reach here, no triplet found
    return false;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 3, 2, 4, 6, 5 };
    int n = arr.length;
    if (checkTriplet(arr, n))
      System.out.print("Yes");
    else
      System.out.print("No");
  }
}
 
// This code is contributed by umadevi9616

C＃

using System;
using System.Collections.Generic;
 
public class GFG {
 
  // Returns true if there is Pythagorean triplet in
  // ar[0..n-1]
  static bool checkTriplet(int []arr, int n) {
 
    // initializing unordered map with key and value as
    // integers
    Dictionary umap = new Dictionary();
 
    // Increase the count of array elements in unordered map
    for (int i = 0; i < n; i++)
      if (umap.ContainsKey(arr[i]))
        umap.Add(arr[i], umap[arr[i]] + 1);
    else
      umap.Add(arr[i], 1);
 
    for (int i = 0; i < n - 1; i++) {
      for (int j = i + 1; j < n; j++) {
 
        // calculating the squares of two elements as
        // integer and float
        int p = (int) Math.Sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
        float q = (float) Math.Sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
 
        // Condition is true if the value is same in
        // integer and float and also the value is
        // present in unordered map
        if (p == q && umap[p] != 0)
          return true;
      }
    }
 
    // If we reach here, no triplet found
    return false;
  }
 
  // Driver Code
  public static void Main(String[] args) {
    int []arr = { 3, 2, 4, 6, 5 };
    int n = arr.Length;
    if (checkTriplet(arr, n))
      Console.Write("Yes");
    else
      Console.Write("No");
  }
}
 
// This code is contributed by umadevi9616

Javascript

 
    // Returns true if there is Pythagorean triplet in
    // ar[0..n-1]
    function checkTriplet(arr , n) {
 
        // initializing unordered map with key and value as
        // integers
        var umap = new Map();
 
        // Increase the count of array elements in unordered map
        for (i = 0; i < n; i++)
            if (umap.has(arr[i]))
                umap.set(arr[i], umap.get(arr[i]) + 1);
            else
                umap.set(arr[i], 1);
 
        for (i = 0; i < n - 1; i++) {
            for (j = i + 1; j < n; j++) {
 
                // calculating the squares of two elements as
                // integer and float
                var p = parseInt( Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]));
                var q =  Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]);
 
                // Condition is true if the value is same in
                // integer and var and also the value is
                // present in unordered map
                if (p == q && umap.get(p) != 0)
                    return true;
            }
        }
 
        // If we reach here, no triplet found
        return false;
    }
 
    // Driver Code
     
        var arr = [ 3, 2, 4, 6, 5 ];
        var n = arr.length;
        if (checkTriplet(arr, n))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by gauravrajput1

输出

Yes

时间复杂度：O(n ² )

方法 5 – 一种更好的基于散列的方法

这种方法使用 Set。首先，我们将对数组的元素进行平方，然后按升序对数组进行排序。运行两个循环，其中外部循环从数组的最后一个索引开始到第二个索引（假设基于 0 的索引），内部循环从 outerLoopIndex – 1 开始到开始。创建一个集合来存储outerLoopIndex 和innerLoopIndex 之间的元素。检查集合中是否存在等于 arr[outerLoopIndex] – arr[innerLoopIndex] 的数字。如果是，则返回“True”。

Python3

# Python program to check if there exists a pythagorean triplet
def checkTriplet(arr, n):
    for i in range(n):
        arr[i] = arr[i] * arr[i]
 
    arr.sort()
 
    for i in range(n - 1, 1, -1):
        s = set()
        for j in range(i - 1, -1, -1):
            if (arr[i] - arr[j]) in s:
                return True
            s.add(arr[j])
    return False
 
# Driver Program
arr = [3, 2, 4, 6, 5]
n = len(arr)
if (checkTriplet(arr, n)):
    print("Yes")
else:
    print("No")
 
# This is contributed by Manvi Pandey

输出

Yes

时间复杂度 – O(n^2)