2024-03-16 02:13:41 +08:00
|
|
|
|
/*
|
|
|
|
|
* File: subset_sum_i_naive.rs
|
|
|
|
|
* Created Time: 2023-07-09
|
|
|
|
|
* Author: codingonion (coderonion@gmail.com)
|
|
|
|
|
*/
|
|
|
|
|
|
|
|
|
|
/* 回溯算法:子集和 I */
|
|
|
|
|
fn backtrack(
|
|
|
|
|
mut state: Vec<i32>,
|
|
|
|
|
target: i32,
|
|
|
|
|
total: i32,
|
|
|
|
|
choices: &[i32],
|
|
|
|
|
res: &mut Vec<Vec<i32>>,
|
|
|
|
|
) {
|
|
|
|
|
// 子集和等于 target 时,记录解
|
|
|
|
|
if total == target {
|
|
|
|
|
res.push(state);
|
|
|
|
|
return;
|
|
|
|
|
}
|
|
|
|
|
// 遍历所有选择
|
|
|
|
|
for i in 0..choices.len() {
|
|
|
|
|
// 剪枝:若子集和超过 target ,则跳过该选择
|
|
|
|
|
if total + choices[i] > target {
|
|
|
|
|
continue;
|
|
|
|
|
}
|
|
|
|
|
// 尝试:做出选择,更新元素和 total
|
|
|
|
|
state.push(choices[i]);
|
|
|
|
|
// 进行下一轮选择
|
|
|
|
|
backtrack(state.clone(), target, total + choices[i], choices, res);
|
|
|
|
|
// 回退:撤销选择,恢复到之前的状态
|
|
|
|
|
state.pop();
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 求解子集和 I(包含重复子集) */
|
|
|
|
|
fn subset_sum_i_naive(nums: &[i32], target: i32) -> Vec<Vec<i32>> {
|
|
|
|
|
let state = Vec::new(); // 状态(子集)
|
|
|
|
|
let total = 0; // 子集和
|
|
|
|
|
let mut res = Vec::new(); // 结果列表(子集列表)
|
|
|
|
|
backtrack(state, target, total, nums, &mut res);
|
|
|
|
|
res
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
pub fn main() {
|
|
|
|
|
let nums = [3, 4, 5];
|
|
|
|
|
let target = 9;
|
|
|
|
|
|
|
|
|
|
let res = subset_sum_i_naive(&nums, target);
|
|
|
|
|
|
|
|
|
|
println!("输入数组 nums = {:?}, target = {}", &nums, target);
|
|
|
|
|
println!("所有和等于 {} 的子集 res = {:?}", target, &res);
|
|
|
|
|
println!("请注意,该方法输出的结果包含重复集合");
|
|
|
|
|
}
|