算法 - 圆形阵列旋转
Algorithms - Circular Array Rotation
圆形阵列旋转的线性复杂性实现是否正确?
n =元素数量k =旋转数
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;
}
no。
考虑此示例。
#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
使用stl ::在std :: array上旋转,您可以旋转,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
屈服:3 4 5 6 1 2,或右键,例如2,为:
std::rotate(begin(a), end(a) - 2, end(a)); // right rotate by 2
屈服:5 6 1 2 3 4,具有线性复杂性。
在left
或right
方向上旋转n
的长度CC_1的数组。
代码在Java中
我定义了一个方向枚举:
public enum Direction {
L, R
};
用times
和direction
旋转:
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)。
示例:输入:[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
旋转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]
旋转3
次right
输出:[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]