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50 lines
1.6 KiB
50 lines
1.6 KiB
/*
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* File: subset_sum_i.rs
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* Created Time: 2023-07-09
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* Author: codingonion (coderonion@gmail.com)
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*/
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/* 回溯算法：子集和 I */
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fn backtrack(mut state: Vec<i32>, target: i32, choices: &[i32], start: usize, res: &mut Vec<Vec<i32>>) {
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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.push(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 i in start..choices.len() {
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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.push(choices[i]);
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// 进行下一轮选择
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backtrack(state.clone(), 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.pop();
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}
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}
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/* 求解子集和 I */
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fn subset_sum_i(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
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let state = Vec::new(); // 状态（子集）
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nums.sort(); // 对 nums 进行排序
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let start = 0; // 遍历起始点
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let mut res = Vec::new(); // 结果列表（子集列表）
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backtrack(state, target, nums, start, &mut res);
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res
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}
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/* Driver Code */
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pub fn main() {
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let mut nums = [ 3, 4, 5 ];
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let target = 9;
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let res = subset_sum_i(&mut nums, target);
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println!("输入数组 nums = {:?}, target = {}", &nums, target);
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println!("所有和等于 {} 的子集 res = {:?}", target, &res);
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} |