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/**
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* File: subset_sum_i.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, choices: [Int], start: Int, res: inout [[Int]]) {
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// 子集和等于 target 时,记录解
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if target == 0 {
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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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// 剪枝二:从 start 开始遍历,避免生成重复子集
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for i in choices.indices.dropFirst(start) {
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// 剪枝一:若子集和超过 target ,则直接结束循环
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// 这是因为数组已排序,后边元素更大,子集和一定超过 target
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if target - choices[i] < 0 {
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break
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}
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// 尝试:做出选择,更新 target, start
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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 - choices[i], choices: choices, start: i, 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 */
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func subsetSumI(nums: [Int], target: Int) -> [[Int]] {
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var state: [Int] = [] // 状态(子集)
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let nums = nums.sorted() // 对 nums 进行排序
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let start = 0 // 遍历起始点
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var res: [[Int]] = [] // 结果列表(子集列表)
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backtrack(state: &state, target: target, choices: nums, start: start, res: &res)
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return res
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}
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@main
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enum SubsetSumI {
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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 = subsetSumI(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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}
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}
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