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题目描述：

解决思路：

• 时间复杂度很低，除了数组排序那块需要O(NlogN)（或者O(MlogM)，看谁最大）之外，后面的顶多需要O(k^2)，应该不是很大。
• 然后其实有O(N)的解法，链接如下：https://blog.csdn.net/huntzw/article/details/25157743

import java.util.*;public class Main {//2,4,2,7,7-3,2,5,6,1,9:6//16 16 13 13 13 11public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String line = scanner.nextLine();String[] tem = line.split("-");String[] array1 = tem[0].split(",");String[] temp = tem[1].split(":");String[] array2 = temp[0].split(",");int K = Integer.parseInt(temp[1]);Arrays.sort(array1, new Comparator<String>() {@Overridepublic int compare(String o1, String o2) {return o2.compareTo(o1);}});Arrays.sort(array2, new Comparator<String>() {@Overridepublic int compare(String o1, String o2) {return o2.compareTo(o1);}});int[] arr_1 = new int[array1.length];int[] arr_2 = new int[array2.length];for(int i = 0; i < array1.length; i++){arr_1[i] = Integer.parseInt(array1[i]);}for(int i = 0; i < array2.length; i++){arr_2[i] = Integer.parseInt(array2[i]);}ArrayList<Integer> sum = new ArrayList<>();int num_1 = K > arr_1.length ? arr_1.length : K;int num_2 = K > arr_2.length ? arr_2.length : K;for(int i = 0; i < num_1; i++){for(int j = 0; j < num_2; j++){sum.add(arr_1[i] + arr_2[j]);}}Collections.sort(sum, new Comparator<Integer>() {@Overridepublic int compare(Integer o1, Integer o2) {return o2 - o1;}});for(int i = 0; i < K; i++)System.out.println(sum.get(i));}
}