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Swift

/**
* File: subset_sum_i_naive.swift
* Created Time: 2023-07-02
* Author: nuomi1 (nuomi1@qq.com)
*/
/* 回溯算法:子集和 I */
func backtrack(state: inout [Int], target: Int, total: Int, choices: [Int], res: inout [[Int]]) {
// 子集和等于 target 时,记录解
if total == target {
res.append(state)
return
}
// 遍历所有选择
for i in stride(from: 0, to: choices.count, by: 1) {
// 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.append(choices[i])
// 进行下一轮选择
backtrack(state: &state, target: target, total: total + choices[i], choices: choices, res: &res)
// 回退:撤销选择,恢复到之前的状态
state.removeLast()
}
}
/* 求解子集和 I(包含重复子集) */
func subsetSumINaive(nums: [Int], target: Int) -> [[Int]] {
var state: [Int] = [] // 状态(子集)
let total = 0 // 子集和
var res: [[Int]] = [] // 结果列表(子集列表)
backtrack(state: &state, target: target, total: total, choices: nums, res: &res)
return res
}
@main
enum SubsetSumINaive {
/* Driver Code */
static func main() {
let nums = [3, 4, 5]
let target = 9
let res = subsetSumINaive(nums: nums, target: target)
print("输入数组 nums = \(nums), target = \(target)")
print("所有和等于 \(target) 的子集 res = \(res)")
print("请注意,该方法输出的结果包含重复集合")
}
}