给定一个数组A []和一个数字x，请检查A []中是否有对，且总和为x |。套装2

给定由N个整数和整数X组成的数组arr [] ，任务是从数组arr []中找到两个具有总和X的元素。如果不存在这样的数字，则打印“ -1” 。

例子：

Input: arr[] = {0, -1, 2, -3, 1}, X = -2
Output: -3, 1
Explanation:
From the given array the sum of -3 and 1 is equal to -2 (= X).

Input: arr[] = {1, -2, 1, 0, 5}, X = 0
Output: -1

方法：可以通过使用排序和二进制搜索来解决给定的问题，其思想是对数组A []进行排序，并针对每个数组元素A [i] ，搜索是否存在另一个值(X – A [i])是否在数组中。请按照以下步骤解决问题：

• 按给定数组arr []的升序排序。
• 遍历数组arr []并为每个数组元素A [i]初始化两个变量low和high分别为0和(N – 1) 。现在，按照以下步骤执行二进制搜索：
  • 如果数组arr []中索引中间的值为(X – A [i]) ，则打印此当前对并退出循环。
  • 更新作为中期(低+高)/ 2。
  • 如果A [MID]的值小于X，然后更新低至(1中间+)。否则，更新高(中- 1)。
• 完成上述步骤后，如果找不到这样的对，则打印-1 。

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 the array has
// 2 elements whose sum is equal to
// the given value
void hasArrayTwoPairs(int nums[],
                      int n, int target)
{
    // Sort the array in increasing order
    sort(nums, nums + n);
 
    // Traverse the array, nums[]
    for (int i = 0; i < n; i++) {
 
        // Store the required number
        // to be found
        int x = target - nums[i];
 
        // Perform binary search
        int low = 0, high = n - 1;
        while (low <= high) {
 
            // Store the mid value
            int mid = low
                      + ((high - low) / 2);
 
            // If nums[mid] is greater
            // than x, then update
            // high to mid - 1
            if (nums[mid] > x) {
                high = mid - 1;
            }
 
            // If nums[mid] is less
            // than x, then update
            // low to mid + 1
            else if (nums[mid] < x) {
                low = mid + 1;
            }
 
            // Otherwise
            else {
 
                // If mid is equal i,
                // check mid-1 and mid+1
                if (mid == i) {
                    if ((mid - 1 >= 0)
                        && nums[mid - 1] == x) {
                        cout << nums[i] << ", ";
                        cout << nums[mid - 1];
                        return;
                    }
                    if ((mid + 1 < n)
                        && nums[mid + 1] == x) {
                        cout << nums[i] << ", ";
                        cout << nums[mid + 1];
                        return;
                    }
                    break;
                }
 
                // Otherwise, print the
                // pair and return
                else {
                    cout << nums[i] << ", ";
                    cout << nums[mid];
                    return;
                }
            }
        }
    }
 
    // If no such pair is found,
    // then print -1
    cout << -1;
}
 
// Driver Code
int main()
{
    int A[] = { 0, -1, 2, -3, 1 };
    int X = -2;
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function Call
    hasArrayTwoPairs(A, N, X);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
 
  // Function to check if the array has
  // 2 elements whose sum is equal to
  // the given value
  static void hasArrayTwoPairs(int nums[],
                               int n, int target)
  {
     
    // Sort the array in increasing order
    Arrays.sort(nums);
 
    // Traverse the array, nums[]
    for (int i = 0; i < n; i++) {
 
      // Store the required number
      // to be found
      int x = target - nums[i];
 
      // Perform binary search
      int low = 0, high = n - 1;
      while (low <= high) {
 
        // Store the mid value
        int mid = low
          + ((high - low) / 2);
 
        // If nums[mid] is greater
        // than x, then update
        // high to mid - 1
        if (nums[mid] > x) {
          high = mid - 1;
        }
 
        // If nums[mid] is less
        // than x, then update
        // low to mid + 1
        else if (nums[mid] < x) {
          low = mid + 1;
        }
 
        // Otherwise
        else {
 
          // If mid is equal i,
          // check mid-1 and mid+1
          if (mid == i) {
            if ((mid - 1 >= 0)
                && nums[mid - 1] == x) {
              System.out.print(nums[i] + ", ");
              System.out.print( nums[mid - 1]);
              return;
            }
            if ((mid + 1 < n)
                && nums[mid + 1] == x) {
              System.out.print( nums[i] + ", ");
              System.out.print( nums[mid + 1]);
              return;
            }
            break;
          }
 
          // Otherwise, print the
          // pair and return
          else {
            System.out.print( nums[i] + ", ");
            System.out.print(nums[mid]);
            return;
          }
        }
      }
    }
 
    // If no such pair is found,
    // then print -1
    System.out.print(-1);
  }
 
 
 
  // Driver Code
  public static void main(String[] args)
  {
    int A[] = { 0, -1, 2, -3, 1 };
    int X = -2;
    int N = A.length;
 
    // Function Call
    hasArrayTwoPairs(A, N, X);
  }
}
 
// This code is contributed by code_hunt.

Python3

# Python3 program for the above approach
 
# Function to check if the array has
# 2 elements whose sum is equal to
# the given value
def hasArrayTwoPairs(nums, n, target):
   
    # Sort the array in increasing order
    nums = sorted(nums)
 
    # Traverse the array, nums[]
    for i in range(n):
 
        # Store the required number
        # to be found
        x = target - nums[i]
 
        # Perform binary search
        low, high = 0, n - 1
        while (low <= high):
 
            # Store the mid value
            mid = low + ((high - low) // 2)
 
            # If nums[mid] is greater
            # than x, then update
            # high to mid - 1
            if (nums[mid] > x):
                high = mid - 1
 
            # If nums[mid] is less
            # than x, then update
            # low to mid + 1
            elif (nums[mid] < x):
                low = mid + 1
 
            # Otherwise
            else:
 
                # If mid is equal i,
                # check mid-1 and mid+1
                if (mid == i):
                    if ((mid - 1 >= 0) and nums[mid - 1] == x):
                        print(nums[i], end = ", ")
                        print(nums[mid - 1])
                        return
                    if ((mid + 1 < n) and nums[mid + 1] == x):
                        print(nums[i], end = ", ")
                        print(nums[mid + 1])
                        return
                    break
                     
                # Otherwise, prthe
                # pair and return
                else:
                    print(nums[i], end = ", ")
                    print(nums[mid])
                    return
 
    # If no such pair is found,
    # then pr-1
    print (-1)
 
# Driver Code
if __name__ == '__main__':
 
    A = [0, -1, 2, -3, 1]
    X = -2
    N = len(A)
 
    # Function Call
    hasArrayTwoPairs(A, N, X)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
public class GFG
{
     
    // Function to check if the array has
    // 2 elements whose sum is equal to
    // the given value
    static void hasArrayTwoPairs(int[] nums, int n, int target)
    {
       
        // Sort the array in increasing order
        Array.Sort(nums);
         
        // Traverse the array, nums[]
        for (int i = 0; i < n; i++)
        {
           
            // Store the required number
            // to be found
            int x = target - nums[i];
  
            // Perform binary search
            int low = 0, high = n - 1;
            while (low <= high)
            {
  
                // Store the mid value
                int mid = low + ((high - low) / 2);
  
                // If nums[mid] is greater
                // than x, then update
                // high to mid - 1
                if (nums[mid] > x) {
                    high = mid - 1;
                }
  
                // If nums[mid] is less
                // than x, then update
                // low to mid + 1
                else if (nums[mid] < x) {
                    low = mid + 1;
                }
  
                // Otherwise
                else {
  
                    // If mid is equal i,
                    // check mid-1 and mid+1
                    if (mid == i) {
                        if ((mid - 1 >= 0) && nums[mid - 1] == x) {
                            Console.Write(nums[i] + ", ");
                            Console.Write( nums[mid - 1]);
                            return;
                        }
                        if ((mid + 1 < n) && nums[mid + 1] == x) {
                            Console.Write( nums[i] + ", ");
                            Console.Write( nums[mid + 1]);
                            return;
                        }
                        break;
                    }
  
                    // Otherwise, print the
                    // pair and return
                    else {
                        Console.Write( nums[i] + ", ");
                        Console.Write(nums[mid]);
                        return;
                    }
                }
            }
        }
  
        // If no such pair is found,
        // then print -1
        Console.Write(-1);
    }
     
    // Driver Code
    static public void Main (){
        int[] A = { 0, -1, 2, -3, 1 };
        int X = -2;
        int N = A.Length;
  
        // Function Call
        hasArrayTwoPairs(A, N, X);
    }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

 
// Javascript program for the above approach
 
// Function to check if the array has
// 2 elements whose sum is equal to
// the given value
function hasArrayTwoPairs(nums, n, target)
{
    // Sort the array in increasing order
    nums.sort();
    var i;
    // Traverse the array, nums[]
    for (i = 0; i < n; i++) {
 
        // Store the required number
        // to be found
        var x = target - nums[i];
 
        // Perform binary search
        var low = 0, high = n - 1;
        while (low <= high) {
 
            // Store the mid value
            var mid = low
                      + (Math.floor((high - low) / 2));
 
            // If nums[mid] is greater
            // than x, then update
            // high to mid - 1
            if (nums[mid] > x) {
                high = mid - 1;
            }
 
            // If nums[mid] is less
            // than x, then update
            // low to mid + 1
            else if (nums[mid] < x) {
                low = mid + 1;
            }
 
            // Otherwise
            else {
 
                // If mid is equal i,
                // check mid-1 and mid+1
                if (mid == i) {
                    if ((mid - 1 >= 0)
                        && nums[mid - 1] == x) {
                        document.write(nums[i] + ", ");
                        document.write(nums[mid - 1]);
                        return;
                    }
                    if ((mid + 1 < n) && nums[mid + 1] == x) {
                        document.write(nums[i] + ", ");
                        document.write(nums[mid + 1]);
                        return;
                    }
                    break;
                }
 
                // Otherwise, print the
                // pair and return
                else {
                    document.write(nums[i] +", ");
                    document.write(nums[mid]);
                    return;
                }
            }
        }
    }
 
    // If no such pair is found,
    // then print -1
    document.write(-1);
}
 
// Driver Code
    var A =  [0, -1, 2, -3, 1];
    var X = -2;
    var N = A.length;
 
    // Function Call
    hasArrayTwoPairs(A, N, X);
 

输出：

-3, 1

时间复杂度： O(N * log N)
辅助空间： O(1)

替代方法：请参阅本文的上一篇文章，以了解更多解决此问题的方法。