swift - 使用回溯的最佳权重子集总和

Question

我正在尝试解决一个问题。我有一些重量。 [2,7,20,70,200,700] 在给定输入后，例如 1507，它应该返回这些权重的最佳组合。在本例中为 [700,200,200,200,200,7]。不幸的是，我的算法返回了 [700, 700, 70, 20, 7, 2, 2, 2, 2, 2]。当我说最佳时，我的意思是我的算法应该使用尽可能少的权重。

func solve(_ targetValue: Int, weights: inout [Int]) -> [Int] {
    // The used weights to store
    var usedWeights: [Int] = []
    // The current total value for the calculation
    var total = targetValue
    // If target value is 0 add it to the array and just return it
    if targetValue == 0 { usedWeights.append(0); return usedWeights }
    // Loop through the weights
    for weight in weights.reversed() {

    while weight <= total {
            total -= weight
            usedWeights.append(weight)
        }
    }    // If still weights are not found call the function recursively again but remove the last element before
    if total != 0 {
        weights.removeLast()
        return solve(targetValue, weights: &weights)
    }
    return usedWeights
}

var newWeights: [Int] = [2,7,20,70,200,700]
print(solve(1507, weights: &newWeights))

我该如何解决这个问题？我究竟做错了什么？重要的是使用回溯来解决它。非常感谢您的帮助。

Answer 1

这是一个可能的递归解决方案:

// Find the shortest combination of (possibly repeating) numbers in `values`
// whose sum is exactly `target`, and whose count is less than `limit`.
// Return `nil` if no such combination exist.
//
// `target` must be non-negative, and `values` an array of positive
// numbers in decreasing order.
//
func solveHelper(target: Int, values: ArraySlice<Int>, limit: Int) -> [Int]? {
    if target == 0 {
        return [] // Target reached exactly.
    }
    guard let first = values.first else {
        return nil // No values left, target cannot be reached.
    }
    if target/first >= limit {
        return nil // Target cannot be reached with less than `limit` values.
    }

    var best: [Int]? = nil   // Best solution found so far
    var bestCount = limit // Number of values in best solution

    for n in stride(from: target/first, through: 0, by: -1) {
        if let s = solveHelper(target: target - n * first, values: values.dropFirst(), limit: bestCount - n) {
            best = s + repeatElement(first, count: n)
            bestCount = s.count + n
        }
    }

    return best
}

// Find the shortest combination of (possibly repeating) values in `values`
// whose sum is exactly `target`. Return `nil` if no such combination exist.
//
// `target` must be non-negative, and `values` an array of positive
// numbers.
//
func solve(target: Int, values: [Int]) -> [Int]? {
    return solveHelper(target: target, values: ArraySlice(values.sorted(by: >)), limit: Int.max)
}

例子:

print(solve(target: 1507, values: [2, 7, 20, 70, 200, 700]) as Any)
// Optional([7, 200, 200, 200, 200, 700])

print(solve(target: 1507, values: [20, 70, 200, 700]) as Any)
// nil

print(solve(target: 6, values: [1, 3, 4]) as Any)
// Optional([3, 3])

print(solve(target: 0, values: [1, 3, 4]) as Any)
// Optional([])

一些解释:

• 假定 target 是非负的，并且所有 values 是积极的。
• solve 对数组进行降序排序，并将其作为一个 ArraySlice 到递归辅助函数。这有助于避免 元素存储的进一步副本，当 values.dropFirst() 传递给递归调用。
• solveHelper 以最大可能数量开始“贪婪” 第一个(即最大的)值，递归地调用自己 剩余的目标总和和值，然后用更少的值重复该过程 第一个值的副本，跟踪找到的最短解决方案 到目前为止。
• 为了“修剪”递归树，传递了一个limit 到递归调用。例如，如果 1507 = 700 + 200 + 200 + 200 + 200 + 7 已经找到，则不再需要仅对 [2, 7, 20, 70]，那只会给出更长的解决方案。
• 该函数在我对给定数组的测试中运行得相当快。 对于更多的可能值，您可能需要更多 复杂的算法，例如动态规划方法 在 Change-making problem 中描述.