C# 程序从两个排序数组中查找最近的对

给定两个排序数组和一个数字 x，找到总和最接近 x 的对，并且该对在每个数组中都有一个元素。

给定两个数组 ar1[0…m-1] 和 ar2[0..n-1] 以及一个数字 x，我们需要找到对 ar1[i] + ar2[j] 使得 (ar1 [i] + ar2[j] – x) 是最小值。

例子：

Input:  ar1[] = {1, 4, 5, 7};
        ar2[] = {10, 20, 30, 40};
        x = 32      
Output:  1 and 30

Input:  ar1[] = {1, 4, 5, 7};
        ar2[] = {10, 20, 30, 40};
        x = 50      
Output:  7 and 40

C#

// C# program to find closest pair in
// an array
using System;
  
class GFG {
  
    // ar1[0..m-1] and ar2[0..n-1] are two
    // given sorted arrays/ and x is given
    // number. This function prints the
    // pair from both arrays such that the
    // sum of the pair is closest to x.
    static void printClosest(int[] ar1,
                             int[] ar2, int m, int n, int x)
    {
  
        // Initialize the diff between pair
        // sum and x.
        int diff = int.MaxValue;
  
        // res_l and res_r are result
        // indexes from ar1[] and ar2[]
        // respectively
        int res_l = 0, res_r = 0;
  
        // Start from left side of ar1[]
        // and right side of ar2[]
        int l = 0, r = n - 1;
        while (l < m && r >= 0) {
  
            // If this pair is closer to
            // x than the previously
            // found closest, then update
            // res_l, res_r and diff
            if (Math.Abs(ar1[l] + ar2[r] - x) < diff) {
                res_l = l;
                res_r = r;
                diff = Math.Abs(ar1[l]
                                + ar2[r] - x);
            }
  
            // If sum of this pair is more
            // than x, move to smaller
            // side
            if (ar1[l] + ar2[r] > x)
                r--;
            else // move to the greater side
                l++;
        }
  
        // Print the result
        Console.Write("The closest pair is ["
                      + ar1[res_l] + ", "
                      + ar2[res_r] + "]");
    }
  
    // Driver program to test above functions
    public static void Main()
    {
        int[] ar1 = { 1, 4, 5, 7 };
        int[] ar2 = { 10, 20, 30, 40 };
        int m = ar1.Length;
        int n = ar2.Length;
        int x = 38;
  
        printClosest(ar1, ar2, m, n, x);
    }
}
  
// This code is contributed by nitin mittal.

输出：

The closest pair is [7, 30]

有关更多详细信息，请参阅有关从两个排序数组中查找最近对的完整文章！