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/*
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* File: subset_sum_i_naive.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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mut state: Vec<i32>,
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target: i32,
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total: i32,
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choices: &[i32],
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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 total == target {
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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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for i in 0..choices.len() {
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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.push(choices[i]);
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// 进行下一轮选择
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backtrack(state.clone(), 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.pop();
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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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fn subset_sum_i_naive(nums: &[i32], target: i32) -> Vec<Vec<i32>> {
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let state = Vec::new(); // 状态(子集)
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let total = 0; // 子集和
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let mut res = Vec::new(); // 结果列表(子集列表)
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backtrack(state, target, total, nums, &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 nums = [3, 4, 5];
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let target = 9;
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let res = subset_sum_i_naive(&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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println!("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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