找到一个三元组，使得两个和等于第三元素

给定一个整数数组，您必须找到三个数字，以使两个元素的总和等于第三个元素。

例子：

Input : {5, 32, 1, 7, 10, 50, 19, 21, 2}
Output : 21, 2, 19

Input : {5, 32, 1, 7, 10, 50, 19, 21, 0}
Output : no such triplet exist

问题来源：Arcesium面试经验|第七组(在校园实习)

简单方法：运行三个循环并检查是否存在三元组，以使两个元素的总和等于第三个元素。

时间复杂度：O(n ^ 3)

高效的方法：这个想法类似于“查找一个三元组，将其总和成给定值”。

• 首先对给定的数组进行排序。
• 从后面开始固定三个中的最大元素，并遍历数组以找到其他两个数字，它们合计为第三个元素。
• 取两个指针j(从前面开始)和k(最初为i-1)以找到两个数字中的最小值，并从i-1中找到两个剩余数字中的最大值
• 如果两个数字的和仍小于A [i]，则需要增加两个数字之和的值，从而增加j指针，从而增加A [j] + A [ k] 。
• 如果两个数字的总和大于A [i]，则需要减小两个数字之和的值，从而减小k指针，从而减小A [j] + A [k的总值] 。

下图是上述方法的模拟：

下面是上述方法的实现：

C++

// CPP program to find three numbers
// such that sum of two makes the
// third element in array
#include 
using namespace std;
  
// Utility function for finding
// triplet in array
void findTriplet(int arr[], int n)
{
    // sort the array
    sort(arr, arr + n);
  
    // for every element in arr
    // check if a pair exist(in array) whose
    // sum is equal to arr element
    for (int i = n - 1; i >= 0; i--) {
        int j = 0;
        int k = i - 1;
  
        // Iterate forward and backward to find
        // the other two elements
        while (j < k) {
  
            // If the two elements sum is
            // equal to the third element
            if (arr[i] == arr[j] + arr[k]) {
  
                // pair found
                cout << "numbers are " << arr[i] << " "
                     << arr[j] << " " << arr[k] << endl;
                return;
            }
  
            // If the element is greater than
            // sum of both the elements, then try
            // adding a smaller number to reach the
            // equality
            else if (arr[i] > arr[j] + arr[k])
                j += 1;
  
            // If the element is smaller, then
            // try with a smaller number
            // to reach equality, so decrease K
            else
                k -= 1;
        }
    }
  
    // No such triplet is found in array
    cout << "No such triplet exists";
}
  
// driver program
int main()
{
    int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    findTriplet(arr, n);
    return 0;
}

Java

// Java program to find three numbers
// such that sum of two makes the
// third element in array
import java.util.Arrays;
  
public class GFG {
  
    // utility function for finding
    // triplet in array
    static void findTriplet(int arr[], int n)
    {
        // sort the array
        Arrays.sort(arr);
  
        // for every element in arr
        // check if a pair exist(in array) whose
        // sum is equal to arr element
        for (int i = n - 1; i >= 0; i--) {
            int j = 0;
            int k = i - 1;
            while (j < k) {
                if (arr[i] == arr[j] + arr[k]) {
  
                    // pair found
                    System.out.println("numbers are " + arr[i] + " "
                                       + arr[j] + " " + arr[k]);
  
                    return;
                }
                else if (arr[i] > arr[j] + arr[k])
                    j += 1;
                else
                    k -= 1;
            }
        }
  
        // no such triplet is found in array
        System.out.println("No such triplet exists");
    }
  
    // driver program
    public static void main(String args[])
    {
        int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 };
        int n = arr.length;
        findTriplet(arr, n);
    }
}
// This code is contributed by Sumit Ghosh

Python

# Python program to find three numbers
# such that sum of two makes the
# third element in array
  
# utility function for finding
# triplet in array
def findTriplet(arr, n):
      
    # sort the array
    arr.sort()
   
    # for every element in arr
    # check if a pair exist(in array) whose
    # sum is equal to arr element
    i = n - 1
    while(i >= 0):
        j = 0
        k = i - 1
        while (j < k):
            if (arr[i] == arr[j] + arr[k]):
                 
                # pair found
                print "numbers are ", arr[i], arr[j], arr[k]
                return
            elif (arr[i] > arr[j] + arr[k]):
                j += 1
            else:
                k -= 1
        i -= 1
          
    # no such triplet is found in array
    print "No such triplet exists"
   
# driver program
arr = [ 5, 32, 1, 7, 10, 50, 19, 21, 2 ]
n = len(arr)
findTriplet(arr, n)
  
# This code is contributed by Sachin Bisht

C#

// C# program to find three numbers
// such that sum of two makes the
// third element in array
using System;
  
public class GFG {
  
    // utility function for finding
    // triplet in array
    static void findTriplet(int[] arr, int n)
    {
  
        // sort the array
        Array.Sort(arr);
  
        // for every element in arr
        // check if a pair exist(in
        // array) whose sum is equal
        // to arr element
        for (int i = n - 1; i >= 0; i--) {
            int j = 0;
            int k = i - 1;
            while (j < k) {
                if (arr[i] == arr[j] + arr[k]) {
  
                    // pair found
                    Console.WriteLine("numbers are "
                                      + arr[i] + " " + arr[j]
                                      + " " + arr[k]);
  
                    return;
                }
                else if (arr[i] > arr[j] + arr[k])
                    j += 1;
                else
                    k -= 1;
            }
        }
  
        // no such triplet is found in array
        Console.WriteLine("No such triplet exists");
    }
  
    // driver program
    public static void Main()
    {
        int[] arr = { 5, 32, 1, 7, 10, 50,
                      19, 21, 2 };
        int n = arr.Length;
  
        findTriplet(arr, n);
    }
}
  
// This code is contributed by vt_m.

PHP

= 0; $i--)
    {
        $j = 0;
        $k = $i - 1;
        while ($j < $k) 
        {
            if ($arr[$i] == $arr[$j] + $arr[$k]) 
            {
                  
                // pair found
                echo "numbers are ", $arr[$i], " ", 
                                      $arr[$j], " ", 
                                      $arr[$k];
                return;
            } 
            else if ($arr[$i] > $arr[$j] + 
                                $arr[$k])
                $j += 1;
            else
                $k -= 1;
        }
    }
  
    // no such triplet 
    // is found in array
    echo "No such triplet exists";
}
  
// Driver Code
$arr = array(5, 32, 1, 7, 10, 
             50, 19, 21, 2 );
$n = count($arr);
  
findTriplet($arr, $n);
  
// This code is contributed by anuj_67.
?>

输出：

numbers are 21 2 19

时间复杂度：O(N ^ 2)