一尘不染

在Swift中旋转数组

algorithm

在Swift中探索算法时,如果不使用funcs shiftLeft/ ,就无法在swift中找到阵列旋转的算法shiftRight

C具有这种优美的算法,其时间复杂度为O(N):

/* Function to left rotate arr[] of size n by d */
void leftRotate(int arr[], int d, int n)
{
    rvereseArray(arr, 0, d-1);
    rvereseArray(arr, d, n-1);
    rvereseArray(arr, 0, n-1);
}

/*Function to reverse arr[] from index start to end*/
void rvereseArray(int arr[], int start, int end)
{
    int temp;
    while (start < end)
    {
        temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

我正在努力将其转换为快速:

func rotate(array:[Int], positions:Int, arSize:Int) {

    var a = array
    var p = positions
    var s = arSize

    reverseArray(array: a, start: 0, end: p-1)
    reverseArray(array: a, start: p, end: s-1)
    reverseArray(array: a, start: 0, end: s-1)
}

func reverseArray(array: [Int], start:Int, end:Int) {

    var a = array
    var s = start
    var e = end
    var temp = 0
    while s < e {
        temp = a[s]
        a[s] = a[e]
        a[e] = temp
        s += 1
        e -= 1
    }
}

据我了解,为了迅速,我们需要指定返回类型。如何配置它们而不增加空间(内存)复杂性?(即,不创建新的临时数组)


这个问题与其他问题不同,因为它returns与C相比如何快速工作。


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2020-07-28

共1个答案

一尘不染

编辑更新:

Swift 5或更高版本

extension RangeReplaceableCollection {
    mutating func rotate(positions: Int) {
        let index = self.index(startIndex, offsetBy: positions, limitedBy: endIndex) ?? endIndex
        let slice = self[..<index]
        removeSubrange(..<index)
        insert(contentsOf: slice, at: endIndex)
    }
}

extension RangeReplaceableCollection where Self: BidirectionalCollection {
    mutating func rotate(positions: Int, size: Int) {
        let index = self.index(startIndex, offsetBy: positions, limitedBy: endIndex) ?? endIndex
        let end = self.index(index, offsetBy: size - positions, limitedBy: self.index(before: endIndex)) ?? endIndex
        replaceSubrange(..<index, with: self[..<index].reversed())
        replaceSubrange(index..<end, with: self[index..<end].reversed())
        replaceSubrange(..<end, with: self[..<end].reversed())
    }
}

var test = [1,2,3,4,5,6,7,8,9,10]
test.rotate(positions: 3)   // [4, 5, 6, 7, 8, 9, 10, 1, 2, 3]

var test2 = "1234567890"
test2.rotate(positions: 3)   // "4567890123"
2020-07-28