swift 3算法-密码标志

Question

我试图在Swift 3中解决codility.com上的旗帜问题，下面的代码只得到6%。低分数的原因是b/c --我的代码没有考虑到其他可能的数组，只是问题中给出的数组。必须有一种方法，我可以使用"for循环“，而不是多个"if语句”的底部。如有任何建议或建设性的批评，将不胜感激。谢谢。

import Foundation

var A = [1, 5, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2]

public func solution(_ A : inout [Int]) -> Int {

    let n = A.count

    if n < 3 {  // if there are only two elements in array A, then there are no peaks

        return 0
    }

    var peaks = [Int]()

    for i in 1..<n - 1 {  // loop to find peak indices

        if A[i - 1] < A[i] && A[i] > A[i + 1] {
            peaks.append(i)
        }
    }

    var numFlags = 0

    if peaks[3] - peaks[2] >= peaks.count {
        numFlags += 2
    }

    if peaks[2] - peaks[1] >= peaks.count {
        numFlags += 1
    }

    if peaks[1] - peaks[0] >= peaks.count {
        numFlags += 1
    }

    if peaks[2] - peaks[0] >= peaks.count {
        numFlags += 1
    }

return numFlags

}

print(solution(&A))

Answer 1

这给了93%的分数，在斯威夫特3关于谦逊的旗帜问题。

import Foundation

var A = [1, 5, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2]

public func solution(_ A : inout [Int]) -> Int {

    let n = A.count  // n = number of indices in A

    var peaks: [Int] = []  // create array for peaks

    if n < 3 {  // you need at least three elements to have a peak
            return 0
        }

    for i in 1...n-2 {  //  loop to determine number of peaks and put them into peaks array
            if A[i - 1] < A[i] && A[i] > A[i + 1] {
                    peaks.append(i)
                }
        }

    if peaks.count == 0 {
            return 0
        }

    var totalMarked = 0

    for i in ((0 + 1)...n).reversed() {
            if (i - 1) * i + 2 > n {  // if this condition is met then iterate for next element
                    continue
                }

            var prevPeak = peaks.first

            var marked = 1

            for j in 1..<peaks.count {
                    if peaks[j] - prevPeak! < i {  // if this condition is met then iterate for next element
                            continue
                        }

                    marked += 1
                    prevPeak = peaks[j]

                    if marked >= i {  // if this condition is met, then exit loop
                            break
                        }
                }

            totalMarked = max(totalMarked, marked)  // take the greater of totalMarked and marked
        }

    return totalMarked
}

print(solution(&A))