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// File: subset_sum_ii.go
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// Created Time: 2023-06-24
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// Author: Reanon (793584285@qq.com)
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package chapter_backtracking
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import "sort"
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/* 回溯算法:子集和 II */
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func backtrackSubsetSumII(start, target int, state, choices *[]int, res *[][]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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newState := append([]int{}, *state...)
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*res = append(*res, newState)
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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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// 剪枝三:从 start 开始遍历,避免重复选择同一元素
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for i := start; i < len(*choices); i++ {
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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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// 剪枝四:如果该元素与左边元素相等,说明该搜索分支重复,直接跳过
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if i > start && (*choices)[i] == (*choices)[i-1] {
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continue
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}
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// 尝试:做出选择,更新 target, start
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*state = append(*state, (*choices)[i])
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// 进行下一轮选择
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backtrackSubsetSumII(i+1, target-(*choices)[i], state, 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 = (*state)[:len(*state)-1]
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}
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}
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/* 求解子集和 II */
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func subsetSumII(nums []int, target int) [][]int {
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state := make([]int, 0) // 状态(子集)
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sort.Ints(nums) // 对 nums 进行排序
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start := 0 // 遍历起始点
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res := make([][]int, 0) // 结果列表(子集列表)
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backtrackSubsetSumII(start, target, &state, &nums, &res)
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return res
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}
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