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/**
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* File: subset_sum_i_naive.dart
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* Created Time: 2023-08-10
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* Author: liuyuxin (gvenusleo@gmail.com)
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*/
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/* 回溯演算法:子集和 I */
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void backtrack(
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List<int> state,
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int target,
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int total,
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List<int> choices,
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List<List<int>> res,
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) {
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// 子集和等於 target 時,記錄解
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if (total == target) {
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res.add(List.from(state));
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return;
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}
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// 走訪所有選擇
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for (int i = 0; i < choices.length; i++) {
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// 剪枝:若子集和超過 target ,則跳過該選擇
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if (total + choices[i] > target) {
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continue;
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}
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// 嘗試:做出選擇,更新元素和 total
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state.add(choices[i]);
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// 進行下一輪選擇
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backtrack(state, target, total + choices[i], choices, res);
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// 回退:撤銷選擇,恢復到之前的狀態
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state.removeLast();
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}
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}
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/* 求解子集和 I(包含重複子集) */
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List<List<int>> subsetSumINaive(List<int> nums, int target) {
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List<int> state = []; // 狀態(子集)
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int total = 0; // 元素和
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List<List<int>> res = []; // 結果串列(子集串列)
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backtrack(state, target, total, nums, res);
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return res;
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}
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/* Driver Code */
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void main() {
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List<int> nums = [3, 4, 5];
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int target = 9;
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List<List<int>> res = subsetSumINaive(nums, target);
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print("輸入陣列 nums = $nums, target = $target");
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print("所有和等於 $target 的子集 res = $res");
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print("請注意,該方法輸出的結果包含重複集合");
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}
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