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/*
* 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" ) ;
}