您可以检查每一对数字并找到最小和。
java代码：
public static void findPairWithMinSumBruteForce(int arr[]){ if(arr.length<2) return; // Suppose 1st two element has minimum sum int minimumSum=arr[0]+arr[1]; int pair1stIndex=0; int pair2ndIndex=1; for (int i = 0; i < arr.length; i++) { for (int j = i+1; j < arr.length; j++) { int tempSum=arr[i]+arr[j]; if(Math.abs(tempSum) < Math.abs(minimumSum)) { pair1stIndex=i; pair2ndIndex=j; minimumSum=tempSum; } } } System.out.println(" The pair whose sum is closest to zero : "+arr[pair1stIndex]+" "+ arr[pair2ndIndex]);}解决方案2：
对数组进行排序。* 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* 计算 arr[l] + arr[r] 的总和* 如果 abs (sum) < abs (minSum)，则更新最小和和对。* 如果 sum 小于 0，这意味着如果我们想找到接近 0 的 sum，请执行 r–* 如果 sum 大于 0，这意味着如果我们想找到接近 0 的 sum ，请执行 l++
java代码：
public static void findPairWithMinSum(int arr[]) { // Sort the array, you can use any sorting algorithm to sort it Arrays.sort(arr); int sum=0; int minimumSum = Integer.MAX_VALUE; int n=arr.length; if(n<0) return; // left and right index variables int l = 0, r = n-1; // variables to keep track of the left and right index pair for minimumSum int minLeft = l, minRight = n-1; while(l < r) { sum = arr[l] + arr[r]; if(Math.abs(sum) < Math.abs(minimumSum)) { minimumSum = sum; minLeft = l; minRight = r; } if(sum < 0) l++; else r--; } System.out.println(" The pair whose sum is minimun : "+arr[minLeft]+" "+ arr[minRight]); }时间复杂度：O(NLogN)
查找总和最接近零的对的 Java 程序：
package org.arpit.java2blog;import java.util.Arrays;public class findPairClosestToZero { public static void main(String[] args) { int array[]={1,30,-5,70,-8,20,-40,60}; findPairWithMinSumBruteForce(array); findPairWithMinSum(array); } public static void findPairWithMinSumBruteForce(int arr[]) { if(arr.length<2) return; // Suppose 1st two element has minimum sum int minimumSum=arr[0]+arr[1]; int pair1stIndex=0; int pair2ndIndex=1; for (int i = 0; i < arr.length; i++) { for (int j = i+1; j < arr.length; j++) { int tempSum=arr[i]+arr[j]; if(Math.abs(tempSum) < Math.abs(minimumSum)) { pair1stIndex=i; pair2ndIndex=j; minimumSum=tempSum; } } } System.out.println(" The pair whose sum is closest to zero using brute force method: "+arr[pair1stIndex]+" "+ arr[pair2ndIndex]); } public static void findPairWithMinSum(int arr[]) { // Sort the array, you can use any sorting algorithm to sort it Arrays.sort(arr); int sum=0; int minimumSum = Integer.MAX_VALUE; int n=arr.length; if(n<0) return; // left and right index variables int l = 0, r = n-1; // variables to keep track of the left and right index pair for minimumSum int minLeft = l, minRight = n-1; while(l < r) { sum = arr[l] + arr[r]; if(Math.abs(sum) < Math.abs(minimumSum)) { minimumSum = sum; minLeft = l; minRight = r; } if(sum < 0) l++; else r--; } System.out.println(" The pair whose sum is minimun : "+arr[minLeft]+" "+ arr[minRight]); }}当你运行程序时，你会得到以下输出：
The pair whose sum is closest to zero using brute force method: 1 -5 The pair whose sum is closest to zero : -5 1
