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/**
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* File: subset_sum_i.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
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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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choices: IntArray,
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start: Int,
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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 (target == 0) {
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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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// 剪枝二:从 start 开始遍历,避免生成重复子集
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for (i in start..<choices.size) {
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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.add(choices[i])
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, 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 */
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fun subsetSumI(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
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val state = mutableListOf<Int>() // 状态(子集)
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nums.sort() // 对 nums 进行排序
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val start = 0 // 遍历起始点
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val res = mutableListOf<MutableList<Int>?>() // 结果列表(子集列表)
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backtrack(state, target, nums, start, 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 = subsetSumI(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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}
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