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	* fix: check the rust codes and fix it * Update binary_tree_bfs.rs --------- Co-authored-by: Yudong Jin <krahets@163.com>
		
			
				
	
	
		
			48 lines
		
	
	
		
			1.5 KiB
		
	
	
	
		
			Rust
		
	
	
	
	
	
			
		
		
	
	
			48 lines
		
	
	
		
			1.5 KiB
		
	
	
	
		
			Rust
		
	
	
	
	
	
/*
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 * File: subset_sum_i_naive.rs
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 * Created Time: 2023-07-09
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 * Author: sjinzh (sjinzh@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, total: i32, choices: &[i32], res: &mut Vec<Vec<i32>>) {
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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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} |