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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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// 子集和等于 target 时,记录解
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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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// 遍历所有选择
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for (int i = 0; i < choices.length; i++) {
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// 剪枝:若子集和超过 target ,则跳过该选择
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if (total + choices[i] > target) {
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continue;
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}
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// 尝试:做出选择,更新元素和 total
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state.add(choices[i]);
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// 进行下一轮选择
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backtrack(state, target, total + choices[i], choices, res);
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// 回退:撤销选择,恢复到之前的状态
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state.removeLast();
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}
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}
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/* 求解子集和 I(包含重复子集) */
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List<List<int>> subsetSumINaive(List<int> nums, int target) {
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List<int> state = []; // 状态(子集)
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int total = 0; // 元素和
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List<List<int>> res = []; // 结果列表(子集列表)
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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("输入数组 nums = $nums, target = $target");
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print("所有和等于 $target 的子集 res = $res");
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print("请注意,该方法输出的结果包含重复集合");
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}
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