题解 | #数组分组#
又是一道dfs的经典题。
注意几个特殊条件
1)sum/2不能整除;
2)输入为5 -5,即sum = 0, target = 0;
import java.util.*; public class Main { public static void main(String[] args) { // 将非5 非3 倍数的数字分为两组,使其差值为5和3组的差值 Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int sum3 = 0, sum5 = 0, sum = 0; List<Integer> list = new ArrayList<>(); for (int i = 0; i < n; i++) { int num = sc.nextInt(); sum += num; if (num % 5 == 0) { sum5 += num; continue; } else if (num % 3 == 0) { sum3 += num; continue; } else { list.add(num); } } if (sum % 2 != 0) { System.out.println("false"); return; } int target = sum / 2 - sum5; if (list.size() == 0 && target == 0) { System.out.println("true"); return; } if (list.size() == 1) { if (list.get(0) != target) { System.out.println("false"); } else { System.out.println("true"); } return; } int[] nums = new int[list.size()]; for (int i = 0; i < nums.length; i++) { nums[i] = list.get(i); } // 组合,m个数据边选择x个数据(不可重复取),使其和为target // 再次安排dfs for (int tag = 1; tag <= nums.length; tag++) { List<List<Integer>> arrange = new ArrayList<>(); List<Integer> tmp = new ArrayList<>(); boolean[] visited = new boolean[nums.length]; dfs(arrange, tmp, nums, visited, tag, 0); for (List<Integer> tmpList : arrange) { int sumList = 0; for (Integer integer : tmpList) { sumList += integer; if (sumList == target) { System.out.println("true"); return; } } } } System.out.println("false"); } /** * 因为需要选择x个数求和,因此只需要组合即可,不需要排列 */ public static void dfs(List<List<Integer>> arrange, List<Integer> tmp, int[] nums, boolean[] visited, int tag, int cur) { if (tmp.size() == tag) { arrange.add(new ArrayList<>(tmp)); return; } for (int i = cur; i < nums.length; i++) { if (!visited[i]) { tmp.add(nums[i]); visited[i] = true; dfs(arrange, tmp, nums, visited, tag,i + 1); // 回溯 visited[i] = false; tmp.remove(tmp.size() - 1); } } } }