/** * 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 { /* 回溯演算法:子集和 I */ static void backtrack(List state, int target, int total, int[] choices, List> res) { // 子集和等於 target 時,記錄解 if (total == target) { res.add(new ArrayList<>(state)); return; } // 走訪所有選擇 for (int i = 0; i < choices.length; i++) { // 剪枝:若子集和超過 target ,則跳過該選擇 if (total + choices[i] > target) { continue; } // 嘗試:做出選擇,更新元素和 total state.add(choices[i]); // 進行下一輪選擇 backtrack(state, target, total + choices[i], choices, res); // 回退:撤銷選擇,恢復到之前的狀態 state.remove(state.size() - 1); } } /* 求解子集和 I(包含重複子集) */ static List> subsetSumINaive(int[] nums, int target) { List state = new ArrayList<>(); // 狀態(子集) int total = 0; // 子集和 List> res = new ArrayList<>(); // 結果串列(子集串列) 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("輸入陣列 nums = " + Arrays.toString(nums) + ", target = " + target); System.out.println("所有和等於 " + target + " 的子集 res = " + res); System.out.println("請注意,該方法輸出的結果包含重複集合"); } }