解决方案1：
您可以检查每对数字并找到接近 X 的总和。
Java 代码：
public static void findPairWithClosestToXBruteForce(int arr[],int X){ if(arr.length<2) return; // Suppose 1st two element has minimum diff with X int minimumDiff=Math.abs(arr[0]+arr[1]-X); int pair1stIndex=0; int pair2ndIndex=1; for (int i = 0; i < arr.length; i++) { for (int j = i+1; j < arr.length; j++) { int tempDiff=Math.abs(arr[i]+arr[j]-X); if(tempDiff< minimumDiff) { pair1stIndex=i; pair2ndIndex=j; minimumDiff=tempDiff; } } } System.out.println(" The pair whose sum is closest to X using brute force method: "+arr[pair1stIndex]+" "+ arr[pair2ndIndex]);}解决方案2：
我们将维护两个索引，一个在开头
（l=0），一个在结尾
（r=n-1）* 迭代直到
l < r* 将差异计算为
arr[l] + arr[r]-x* 如果
abs (diff) < minDiff然后更新最小总和和对。* 如果
arr[l] + arr[r]小于 X，这意味着如果我们想找到接近 X 的和，做 r–* 如果
arr[l] + arr[r]大于 0，这意味着如果我们想找到接近 X 的总和，请执行
l++
java代码：
public static void findPairWithClosestToX(int arr[],int X) { int minimumDiff = Integer.MAX_VALUE; int n=arr.length; if(n<0) return; // left and right index variables int l = 0, r = n-1; // variable to keep track of the left and right pair for minimumSum int minLeft = l, minRight = n-1; while(l < r) { int currentDiff= Math.abs(arr[l] + arr[r]-X); if(currentDiff < minimumDiff) { minimumDiff = currentDiff; minLeft = l; minRight = r; } if(arr[l] + arr[r] < X) l++; else r--; } System.out.println(" The pair whose sum is closest to X : "+arr[minLeft]+" "+ arr[minRight]);}时间复杂度：O(NLogN)
寻找总和最接近 X 的对的 Java 程序：
package org.arpit.java2blog;public class FindPairClosestToXMain { public static void main(String[] args) { int array[]={-40,-5,1,3,6,7,8,20}; findPairWithClosestToXBruteForce(array,5); findPairWithClosestToX(array,5); } public static void findPairWithClosestToXBruteForce(int arr[],int X) { if(arr.length<2) return; // Suppose 1st two element has minimum diff with X int minimumDiff=Math.abs(arr[0]+arr[1]-X); int pair1stIndex=0; int pair2ndIndex=1; for (int i = 0; i < arr.length; i++) { for (int j = i+1; j < arr.length; j++) { int tempDiff=Math.abs(arr[i]+arr[j]-X); if(tempDiff< minimumDiff) { pair1stIndex=i; pair2ndIndex=j; minimumDiff=tempDiff; } } } System.out.println(" The pair whose sum is closest to X using brute force method: "+arr[pair1stIndex]+" "+ arr[pair2ndIndex]); } public static void findPairWithClosestToX(int arr[],int X) { int minimumDiff = Integer.MAX_VALUE; int n=arr.length; if(n<0) return; // left and right index variables int l = 0, r = n-1; // variable to keep track of the left and right pair for minimumSum int minLeft = l, minRight = n-1; while(l < r) { int currentDiff= Math.abs(arr[l] + arr[r]-X); if(currentDiff < minimumDiff) { minimumDiff = currentDiff; minLeft = l; minRight = r; } if(arr[l] + arr[r] < X) l++; else r--; } System.out.println(" The pair whose sum is closest to X : "+arr[minLeft]+" "+ arr[minRight]); }}当你运行上面的程序时，你会得到以下输出：
The pair whose sum is closest to X using brute force method: 1 3The pair whose sum is closest to X : 1 3
