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57 lines
1.5 KiB
57 lines
1.5 KiB
/**
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* File: subset_sum_i.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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List<int> choices,
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int start,
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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 (target == 0) {
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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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// 剪枝二：从 start 开始遍历，避免生成重复子集
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for (int i = start; i < choices.length; i++) {
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// 剪枝一：若子集和超过 target ，则直接结束循环
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// 这是因为数组已排序，后边元素更大，子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 尝试：做出选择，更新 target, start
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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 - choices[i], choices, i, 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 */
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List<List<int>> subsetSumI(List<int> nums, int target) {
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List<int> state = []; // 状态（子集）
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nums.sort(); // 对 nums 进行排序
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int start = 0; // 遍历起始点
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List<List<int>> res = []; // 结果列表（子集列表）
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backtrack(state, target, nums, start, 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 = subsetSumI(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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}
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