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			52 lines
		
	
	
		
			1.6 KiB
		
	
	
	
		
			Swift
		
	
	
	
	
	
			
		
		
	
	
			52 lines
		
	
	
		
			1.6 KiB
		
	
	
	
		
			Swift
		
	
	
	
	
	
/**
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 * File: subset_sum_i_naive.swift
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 * Created Time: 2023-07-02
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 * Author: nuomi1 (nuomi1@qq.com)
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 */
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/* 回溯算法:子集和 I */
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func backtrack(state: inout [Int], target: Int, total: Int, choices: [Int], res: inout [[Int]]) {
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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.append(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 stride(from: 0, to: choices.count, by: 1) {
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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.append(choices[i])
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        // 进行下一轮选择
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        backtrack(state: &state, target: target, total: total + choices[i], choices: choices, res: &res)
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        // 回退:撤销选择,恢复到之前的状态
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        state.removeLast()
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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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func subsetSumINaive(nums: [Int], target: Int) -> [[Int]] {
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    var state: [Int] = [] // 状态(子集)
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    let total = 0 // 子集和
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    var res: [[Int]] = [] // 结果列表(子集列表)
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    backtrack(state: &state, target: target, total: total, choices: nums, res: &res)
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    return res
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}
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@main
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enum SubsetSumINaive {
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    /* Driver Code */
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    static func main() {
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        let nums = [3, 4, 5]
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        let target = 9
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        let res = subsetSumINaive(nums: nums, target: target)
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        print("输入数组 nums = \(nums), target = \(target)")
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        print("所有和等于 \(target) 的子集 res = \(res)")
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        print("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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}
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