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/**
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* File: subset_sum_i_native.kt
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* Created Time: 2024-01-25
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* Author: curtishd (1023632660@qq.com)
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*/
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package chapter_backtracking.subset_sum_i_naive
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/* 回溯算法:子集和 I */
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fun backtrack(
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state: MutableList<Int>,
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target: Int,
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total: Int,
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choices: IntArray,
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res: MutableList<MutableList<Int>?>
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) {
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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.add(state.toMutableList())
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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 choices.indices) {
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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.add(choices[i])
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// 进行下一轮选择
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backtrack(state, target, total + choices[i], choices, res)
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// 回退:撤销选择,恢复到之前的状态
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state.removeAt(state.size - 1)
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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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fun subsetSumINaive(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
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val state = mutableListOf<Int>() // 状态(子集)
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val total = 0 // 子集和
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val res = mutableListOf<MutableList<Int>?>() // 结果列表(子集列表)
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backtrack(state, target, total, nums, res)
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return res
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}
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/* Driver Code */
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fun main() {
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val nums = intArrayOf(3, 4, 5)
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val target = 9
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val res = subsetSumINaive(nums, target)
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println("输入数组 nums = ${nums.contentToString()}, target = $target")
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println("所有和等于 $target 的子集 res = $res")
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println("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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