/** * File: subset_sum_i_naive.java * Created Time: 2023-06-21 * Author: krahets (krahets@163.com) */ package chapter_backtracking; import java.util.*; public class subset_sum_i_naive { /* 回溯算法:子集和 I */ static void backtrack(List state, int target, int total, int[] choices, List> res) { // 子集和等于 target 时,记录解 if (total == target) { res.add(new ArrayList<>(state)); return; } // 遍历所有选择 for (int i = 0; i < choices.length; i++) { // 剪枝:若子集和超过 target ,则跳过该选择 if (total + choices[i] > target) { continue; } // 尝试:做出选择,更新元素和 total state.add(choices[i]); // 进行下一轮选择 backtrack(state, target, total + choices[i], choices, res); // 回退:撤销选择,恢复到之前的状态 state.remove(state.size() - 1); } } /* 求解子集和 I(包含重复子集) */ static List> subsetSumINaive(int[] nums, int target) { List state = new ArrayList<>(); // 状态(子集) int total = 0; // 子集和 List> res = new ArrayList<>(); // 结果列表(子集列表) backtrack(state, target, total, nums, res); return res; } public static void main(String[] args) { int[] nums = { 3, 4, 5 }; int target = 9; List> res = subsetSumINaive(nums, target); System.out.println("输入数组 nums = " + Arrays.toString(nums) + ", target = " + target); System.out.println("所有和等于 " + target + " 的子集 res = " + res); System.out.println("请注意,该方法输出的结果包含重复集合"); } }