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@@ -220,9 +220,36 @@ comments: true
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=== "Swift"
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```swift title="subset_sum_i_naive.swift"
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[class]{}-[func]{backtrack}
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/* 回溯算法:子集和 I */
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func backtrack(state: inout [Int], target: Int, total: Int, choices: [Int], res: inout [[Int]]) {
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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.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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for i in stride(from: 0, to: choices.count, by: 1) {
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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.append(choices[i])
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// 进行下一轮选择
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backtrack(state: &state, target: target, total: total + choices[i], choices: choices, 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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[class]{}-[func]{subsetSumINaive}
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/* 求解子集和 I(包含重复子集) */
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func subsetSumINaive(nums: [Int], target: Int) -> [[Int]] {
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var state: [Int] = [] // 状态(子集)
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let total = 0 // 子集和
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var res: [[Int]] = [] // 结果列表(子集列表)
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backtrack(state: &state, target: target, total: total, choices: nums, res: &res)
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return res
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}
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```
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=== "Zig"
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@@ -485,9 +512,39 @@ comments: true
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=== "Swift"
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```swift title="subset_sum_i.swift"
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[class]{}-[func]{backtrack}
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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 stride(from: start, to: choices.count, by: 1) {
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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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[class]{}-[func]{subsetSumI}
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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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```
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=== "Zig"
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@@ -767,9 +824,44 @@ comments: true
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=== "Swift"
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```swift title="subset_sum_ii.swift"
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[class]{}-[func]{backtrack}
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/* 回溯算法:子集和 II */
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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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// 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 in stride(from: start, to: choices.count, by: 1) {
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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(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 + 1, 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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[class]{}-[func]{subsetSumII}
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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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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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```
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=== "Zig"
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