题解 | #数组分组#

当存在 sum5 + x = sum3 + y 时符合题意 x + y = sunOther; sum5 - sum3 + sumOther = 2y; 只要在OtherList中只要找到y 就可以得到解。递归寻找y

import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    static boolean result = false;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNextInt()) { // 注意 while 处理多个 case
            int count = in.nextInt();
            int[] number = new int[count];
            for (int i = 0; i < count; i++) {
                number[i] = in.nextInt();
            }
            System.out.println(findNumnerArray(number));
        }
    }
    public static boolean findNumnerArray(int[] number) {
        int sumOther = 0;
        int sum5 = 0;
        int sum3 = 0;
        List<Integer> list = new ArrayList<>();
        for (int temp : number) {
            if (temp % 5 == 0) {
                sum5 = sum5 + temp;
            } else if (temp % 3 == 0) {
                sum3 = sum3 + temp;
            } else {
                sumOther = sumOther + temp;
                list.add(temp);
            }
        }
        int target = (sum5 - sum3 + sumOther) / 2;
        if ((sum5 + sum3 + sumOther) % 2 != 0) return false;
        dfs(list, target, 0);
        return result;
    }
    public static void dfs (List<Integer> list, int target, int index) {
        if (index == list.size()) {
            if (target == 0) result = true;
            return;
        }
        dfs(list, target - list.get(index), index + 1);
        dfs(list, target, index + 1);
    }
}