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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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/* 回溯演算法: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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// 子集和等於 target 時,記錄解
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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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// 走訪所有選擇
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// 剪枝二:從 start 開始走訪,避免生成重複子集
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for i in choices.indices.dropFirst(start) {
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// 剪枝一:若子集和超過 target ,則直接結束迴圈
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// 這是因為陣列已排序,後邊元素更大,子集和一定超過 target
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if target - choices[i] < 0 {
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break
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}
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// 嘗試:做出選擇,更新 target, start
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state.append(choices[i])
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// 進行下一輪選擇
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backtrack(state: &state, target: target - choices[i], choices: choices, start: i, res: &res)
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// 回退:撤銷選擇,恢復到之前的狀態
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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] = [] // 狀態(子集)
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let nums = nums.sorted() // 對 nums 進行排序
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let start = 0 // 走訪起始點
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var res: [[Int]] = [] // 結果串列(子集串列)
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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("輸入陣列 nums = \(nums), target = \(target)")
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print("所有和等於 \(target) 的子集 res = \(res)")
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}
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}
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