Question

圆形阵列旋转的线性复杂性实现是否正确？

n =元素数量 k =转数

    int write_to     = 0;
    int copy_current = 0;
    int copy_final   = a[0];
    int rotation     = k;
    int position     = 0;

    for (int i = 0; i < n; i++) {
        write_to     = (position + rotation) % n;
        copy_current = a[write_to];
        a[write_to]  = copy_final;
        position     = write_to;
        copy_final   = copy_current;
     }

Answer 1

没有

考虑这个例子。

#include <iostream>

int main(void) {
    int n = 6;
    int k = 2;
    int a[] = {1, 2, 3, 4, 5, 6};

    int write_to     = 0;
    int copy_current = 0;
    int copy_final   = a[0];
    int rotation     = k;
    int position     = 0;

    for (int i = 0; i < n; i++) {
        write_to     = (position + rotation) % n;
        copy_current = a[write_to];
        a[write_to]  = copy_final;
        position     = write_to;
        copy_final   = copy_current;
     }

    for (int i = 0; i < n; i++) {
        std::cout << a[i] << (i + 1 < n ? ' ' : '\n');
    }
    return 0;
}

预期结果：

5 6 1 2 3 4

实际结果：

3 2 1 4 1 6

Answer 2

在std :: array上使用stl :: rotate，你可以将其旋转，例如2，as：

std::array<int, 6> a{1, 2, 3, 4, 5, 6};
std::rotate(begin(a), begin(a) + 2, end(a)); // left rotate by 2

得出：3 4 5 6 1 2，或者右转，比如2，如：

std::rotate(begin(a), end(a) - 2, end(a)); // right rotate by 2

产量：5 6 1 2 3 4，具有线性复杂度。

Answer 3

在n或k方向上left次旋转长度为right的数组。

代码是Java

我定义了一个方向枚举：

public enum Direction {
    L, R
};

使用times和direction进行轮播：

public static final void rotate(int[] arr, int times, Direction direction) {
    if (arr == null || times < 0) {
        throw new IllegalArgumentException("The array must be non-null and the order must be non-negative");
    }
    int offset = arr.length - times % arr.length;
    if (offset > 0) {
        int[] copy = arr.clone();
        for (int i = 0; i < arr.length; ++i) {
            int j = (i + offset) % arr.length;
            if (Direction.R.equals(direction)) {
                arr[i] = copy[j];
            } else {
                arr[j] = copy[i];
            }
        }
    }
}

复杂性：O（n）。

实施例：输入：[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 轮流3次left 输出：[4, 5, 6, 7, 8, 9, 10, 1, 2, 3]

输入：[4, 5, 6, 7, 8, 9, 10, 1, 2, 3] 轮流3次right 输出：[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

算法 - 圆阵旋转

3 个答案: