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8a6ce26f6a
* idomatic rust * More idiomatic rust * make rust code more idiomatic * update
56 lines
1.6 KiB
Rust
56 lines
1.6 KiB
Rust
/*
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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(
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state: &mut Vec<i32>,
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target: i32,
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choices: &[i32],
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start: usize,
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res: &mut Vec<Vec<i32>>,
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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.push(state.clone());
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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, 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 mut 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(&mut 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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}
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