Question

我正在努力解决问题。我有一些重量。 [2,7,20,70,200,700] 在给定输入后，例如1507，它应返回这些权重的最佳组合。在这种情况下，[700,200,200,200,200,7]。不幸的是，我的算法正在返回[700, 700, 70, 20, 7, 2, 2, 2, 2, 2]。当我说最优时，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。我非常感谢你的帮助。

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)
}

示例：

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：

假设target为非负数且所有values 是积极的。
solve按降序对数组进行排序，并将其作为传递给它 ArraySlice到递归辅助函数。这有助于避免 values.dropFirst()时元素存储的更多副本被传递给递归调用。
solveHelper以最大可能的数量开始“贪婪” 第一个（即最大的）值，递归调用自身剩余的目标总和和值，然后用较少的重复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limit 到递归调用。例如，如果已经找到1507 = 700 + 200 + 200 + 200 + 200 + 7，那么就不再需要仅对[2, 7, 20, 70]中的值求和，这只会提供更长的解决方案。
在我对给定数组的测试中，该函数运行速度相当快。对于更多可能的值，您可能需要更多复杂的算法，如动态编程方法在Change-making problem中描述。

使用回溯的最佳权重子集和

1 个答案: