Algorithm M位圆移位N尺寸阵列的最快算法

对于M个位置的圆移位阵列，最快的算法是什么
例如，

[3 4 5 2 3 1 4]

移位M=2位置应为

[1 4 3 4 5 2 3]

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非常感谢。

使用指针进行设置，几乎不需要时间。每个元素都指向下一个，而“last”（没有last；毕竟，您说过它是循环的）指向第一个元素。一个指向“开始”（第一个元素）的指针，可能还有一个长度，就有了数组。现在，要进行轮班，只需沿着圆圈移动起始指针

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要求一个好的算法，你就会得到明智的想法。要求最快的速度，你会得到奇怪的想法

根据您使用的数据结构，您可以在O（1）中完成。我认为最快的方法是以链表的形式保存数组，并拥有一个哈希表，该哈希表可以在数组中的“索引”到条目的“指针”之间进行转换。这样，您可以在O（1）中找到相关的头和尾，并在O（1）中重新连接（并在O（1）中的切换之后更新哈希表）。这当然是一个非常“混乱”的解决方案，但如果您只对移位的速度感兴趣，那就可以了（代价是在数组中插入和查找的时间更长，但仍然是O（1））

如果你有一个纯数组中的数据，我认为你不能避免O（n）

就编码而言，这取决于您使用的语言

例如，在Python中，您可以“切片”它（假设n是移位大小）：

---

（我知道哈希查找在理论上不是O（1），但我们在这里是实际的，而不是理论的，至少我希望如此…

这只是一个表示问题。将当前索引保持为整数变量，并在遍历数组时使用模运算符了解何时换行。然后，移位只是更改当前索引的值，将其环绕在数组的大小上。这当然是O（1）

例如：

int index = 0;
Array a = new Array[SIZE];

get_next_element() {
    index = (index + 1) % SIZE; 
    return a[index];
}

shift(int how_many) {
    index = (index+how_many) % SIZE;
}

如果您想要O（n）时间且不需要额外的内存使用（因为指定了数组），请使用Jon Bentley的书《编程珍珠第二版》中的算法。它将所有元素交换两次。没有使用链表那么快，但是使用更少的内存，而且概念上很简单

shiftArray( theArray, M ):
    size = len( theArray )
    assert( size > M )
    reverseArray( theArray, 0, size - 1 )
    reverseArray( theArray, 0, M - 1 )
    reverseArray( theArray, M, size - 1 )

---

reverseArray（anArray、startIndex、endIndex）将元素的顺序从startIndex反转为endIndex（包括endIndex）。

将两个索引保留到数组，一个索引从数组的开头开始到数组的结尾。另一个索引从最后的第M个位置开始，并在最后的M个元素中循环任意次数。始终取O（n）。不需要额外的空间

circleArray(Elements,M){
 int size=size-of(Elements);

 //first index
 int i1=0;

 assert(size>M)

 //second index starting from mth position from the last
 int i2=size-M;

 //until first index reaches the end
 while(i1<size-1){

  //swap the elements of the array pointed by both indexes
  swap(i1,i2,Elements);

  //increment first pointer by 1
  i1++;

  //increment second pointer. if it goes out of array, come back to
  //mth position from the last
  if(++i2==size) i2=size-M;

 }
}

circleArray（元素，M）{
int size=元素的大小；
//第一索引
int i1=0；
断言（大小>M）
//第二个索引从最后一个位置的第m个位置开始
int i2=尺寸-M；
//直到第一个索引到达末尾
而（i1
circleArray
有一些错误，并且在所有情况下都不工作

当i1

不是

i1

时，循环必须继续

void Shift(int* _array, int _size, int _moves)
{
    _moves = _size - _moves;
    int i2 = _moves;
         int i1 = -1;
         while(++i1 < i2)
    {
        int tmp = _array[i2];
        _array[i2] = _array[i1];
        _array[i1] = tmp;
        if(++i2 == _size) i2 = _moves;
    }
}

void Shift（int*\u数组、int大小、int移动）
{
_移动=\u大小-\u移动；
int i2=_移动；
inti1=-1；
而（++i1
静态int[]移位（int-arr[]，int-index，int-k，int-rem）
{
如果（k=阵列长度）
{
返回arr；
}
int temp=arr[索引]；
arr=位移（arr，（指数+k）%arr.length，k，rem-1）；
arr[（索引+k）%arr.length]=温度；
返回arr；
}
如果您对Java实现感兴趣，请参阅以下内容：

C数组shift right函数。如果shift为负数，则函数将数组向左移动。
它经过优化以减少内存使用。运行时间为O（n）

void arrayshift right（int-array[]，int-size，int-shift）{
内伦；
//减班
移位%=大小；
//如果换档小于0-重定向向左换档
如果（移位<0）{
移位+=大小；
}
len=尺寸-位移；
//选择需要较少内存的算法
如果（移位=0；i--，j--）{
数组[i]=数组[j]；
}
//从tmp数组插入丢失的值
for（int i=0；i

此代码即使在负移位k时也能正常工作这应该可以循环移位arry：
输入：{1,2,3,5,6,7,8}；
forloops之后数组中存在的输出值：{8,7,1,2,3,5,6,8,7}

 class Program
    {
        static void Main(string[] args)
        {
            int[] array = { 1, 2, 3, 5, 6, 7, 8 };
            int index = 2;
            int[] tempArray = new int[array.Length];
            array.CopyTo(tempArray, 0);

            for (int i = 0; i < array.Length - index; i++)
            {
                array[index + i] = tempArray[i];
            }

            for (int i = 0; i < index; i++)
            {
                array[i] = tempArray[array.Length -1 - i];
            }            
        }
    }

类程序
{
静态void Main（字符串[]参数）
{
int[]数组={1,2,3,5,6,7,8}；
int指数=2；
int[]tempArray=newint[array.Length]；
CopyTo（tempArray，0）；
for（int i=0；i
Ruby示例：

def move_cyclic2 array, move_cnt
  move_cnt = array.length - move_cnt % array.length 
  if !(move_cnt == 0 || move_cnt == array.length)            
    array.replace( array[move_cnt..-1] + array[0...move_cnt] )  
  end   
end

理论上，最快的循环是这样的：

if (begin != middle && middle != end)
{
    for (i = middle; ; )
    {
        swap(arr[begin++], arr[i++]);
        if (begin == middle && i == end) { break; }
        if (begin == middle) { middle = i; }
        else if (i == end) { i = middle; }
    }
}

在实践中，你应该专业
 class Program
    {
        static void Main(string[] args)
        {
            int[] array = { 1, 2, 3, 5, 6, 7, 8 };
            int index = 2;
            int[] tempArray = new int[array.Length];
            array.CopyTo(tempArray, 0);

            for (int i = 0; i < array.Length - index; i++)
            {
                array[index + i] = tempArray[i];
            }

            for (int i = 0; i < index; i++)
            {
                array[i] = tempArray[array.Length -1 - i];
            }            
        }
    }

def move_cyclic2 array, move_cnt
  move_cnt = array.length - move_cnt % array.length 
  if !(move_cnt == 0 || move_cnt == array.length)            
    array.replace( array[move_cnt..-1] + array[0...move_cnt] )  
  end   
end

if (begin != middle && middle != end)
{
    for (i = middle; ; )
    {
        swap(arr[begin++], arr[i++]);
        if (begin == middle && i == end) { break; }
        if (begin == middle) { middle = i; }
        else if (i == end) { i = middle; }
    }
}

void shift_vec(vector<int>& v, size_t a)
{
    size_t max_s = v.size() / a;
    for( size_t s = 1; s < max_s; ++s )
        for( size_t i = 0; i < a; ++i )
            swap( v[i], v[s*a+i] );
    for( size_t i = 0; i < a; ++i )
        swap( v[i], v[(max_s*a+i) % v.size()] );
}

var shiftLeft = function(list, m) {
    var from = 0;
    var val = list[from];
    var nextGroup = 1;
    for(var i = 0; i < list.length; i++) {
        var to = ((from - m) + list.length) % list.length;
        if(to == from)
            break;

        var temp = list[to];
        list[to] = val;
        from = to;
        val = temp;

        if(from < nextGroup) {
            from = nextGroup++;
            val = list[from];
        }
    }
    return list;
}

tmp = arr[0]; arr[0] = arr[1]; ... arr[7] = arr[8]; arr[8] = tmp;

// n is length(arr)
// shift is how many place to cycle shift left
void cycle_shift_left(int arr[], int n, int shift) {
  int i, j, k, tmp;
  if(n <= 1 || shift == 0) return;
  shift = shift % n; // make sure shift isn't >n
  int gcd = calc_GCD(n, shift);

  for(i = 0; i < gcd; i++) {
    // start cycle at i
    tmp = arr[i];
    for(j = i; 1; j = k) {
      k = j+shift;
      if(k >= n) k -= n; // wrap around if we go outside array
      if(k == i) break; // end of cycle
      arr[j] = arr[k];
    }
    arr[j] = tmp;
  }
}

// circle shift an array left (towards index zero)
// - ptr    array to shift
// - n      number of elements
// - es     size of elements in bytes
// - shift  number of places to shift left
void array_cycle_left(void *_ptr, size_t n, size_t es, size_t shift)
{
  char *ptr = (char*)_ptr;
  if(n <= 1 || !shift) return; // cannot mod by zero
  shift = shift % n; // shift cannot be greater than n

  // Using GCD
  size_t i, j, k, gcd = calc_GCD(n, shift);
  char tmp[es];

  // i is initial starting position
  // Copy from k -> j, stop if k == i, since arr[i] already overwritten
  for(i = 0; i < gcd; i++) {
    memcpy(tmp, ptr+es*i, es); // tmp = arr[i]
    for(j = i; 1; j = k) {
      k = j+shift;
      if(k >= n) k -= n;
      if(k == i) break;
      memcpy(ptr+es*j, ptr+es*k, es); // arr[j] = arr[k];
    }
    memcpy(ptr+es*j, tmp, es); // arr[j] = tmp;
  }
}

// cycle right shifts away from zero
void array_cycle_right(void *_ptr, size_t n, size_t es, size_t shift)
{
  if(!n || !shift) return; // cannot mod by zero
  shift = shift % n; // shift cannot be greater than n
  // cycle right by `s` is equivalent to cycle left by `n - s`
  array_cycle_left(_ptr, n, es, n - shift);
}

// Get Greatest Common Divisor using binary GCD algorithm
// http://en.wikipedia.org/wiki/Binary_GCD_algorithm
unsigned int calc_GCD(unsigned int a, unsigned int b)
{
  unsigned int shift, tmp;

  if(a == 0) return b;
  if(b == 0) return a;

  // Find power of two divisor
  for(shift = 0; ((a | b) & 1) == 0; shift++) { a >>= 1; b >>= 1; }

  // Remove remaining factors of two from a - they are not common
  while((a & 1) == 0) a >>= 1;

  do
  {
    // Remove remaining factors of two from b - they are not common
    while((b & 1) == 0) b >>= 1;

    if(a > b) { tmp = a; a = b; b = tmp; } // swap a,b
    b = b - a;
  }
  while(b != 0);

  return a << shift;
}

#include <iostream>

#include <assert.h>

#include <cstring>

using namespace std;

struct VeryElaboratedDataType
{
    int a;
    int b;
};

namespace amsoft
{
    namespace inutils
    {
        enum EShiftDirection
        {
            Left,
            Right
        };
template 
<typename T,size_t len>
void infernalShift(T infernalArray[],int positions,EShiftDirection direction = EShiftDirection::Right)
{
    //assert the dudes
    assert(len > 0 && "what dude?");
    assert(positions >= 0 && "what dude?");

    if(positions > 0)
    {
    ++positions;
    //let's make it fit the range
    positions %= len;

    //if y want to live as a forcio, i'l get y change direction by force
    if(!direction)
    {
        positions = len - positions;
    }

    // here I prepare a fine block of raw memory... allocate once per thread
    static unsigned char WORK_BUFFER[len * sizeof(T)];
    // std::memset (WORK_BUFFER,0,len * sizeof(T));
    // clean or not clean?, well
    // Hamlet is a prince, a prince does not clean

    //copy the first chunk of data to the 0 position
    std::memcpy(WORK_BUFFER,reinterpret_cast<unsigned char *>(infernalArray) + (positions)*sizeof(T),(len - positions)*sizeof(T));
    //copy the second chunk of data to the len - positions position
    std::memcpy(WORK_BUFFER+(len - positions)*sizeof(T),reinterpret_cast<unsigned char *>(infernalArray),positions * sizeof(T));

    //now bulk copy back to original one
    std::memcpy(reinterpret_cast<unsigned char *>(infernalArray),WORK_BUFFER,len * sizeof(T));

    }

}
template 
<typename T>
void printArray(T infernalArrayPrintable[],int len)
{
        for(int i=0;i<len;i++)
    {
        std::cout << infernalArrayPrintable[i] << " ";
    }
    std::cout << std::endl;

}
template 
<>
void printArray(VeryElaboratedDataType infernalArrayPrintable[],int len)
{
        for(int i=0;i<len;i++)
    {
        std::cout << infernalArrayPrintable[i].a << "," << infernalArrayPrintable[i].b << " ";
    }
    std::cout << std::endl;

}
}
}




int main() {
    // your code goes here
    int myInfernalArray[] = {1,2,3,4,5,6,7,8,9};

    VeryElaboratedDataType myInfernalArrayV[] = {{1,1},{2,2},{3,3},{4,4},{5,5},{6,6},{7,7},{8,8},{9,9}};
    amsoft::inutils::printArray(myInfernalArray,sizeof(myInfernalArray)/sizeof(int));
    amsoft::inutils::infernalShift<int,sizeof(myInfernalArray)/sizeof(int)>(myInfernalArray,4);
    amsoft::inutils::printArray(myInfernalArray,sizeof(myInfernalArray)/sizeof(int));
    amsoft::inutils::infernalShift<int,sizeof(myInfernalArray)/sizeof(int)>(myInfernalArray,4,amsoft::inutils::EShiftDirection::Left);
    amsoft::inutils::printArray(myInfernalArray,sizeof(myInfernalArray)/sizeof(int));
    amsoft::inutils::infernalShift<int,sizeof(myInfernalArray)/sizeof(int)>(myInfernalArray,10);
    amsoft::inutils::printArray(myInfernalArray,sizeof(myInfernalArray)/sizeof(int));


    amsoft::inutils::printArray(myInfernalArrayV,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType));
    amsoft::inutils::infernalShift<VeryElaboratedDataType,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType)>(myInfernalArrayV,4);
    amsoft::inutils::printArray(myInfernalArrayV,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType));
    amsoft::inutils::infernalShift<VeryElaboratedDataType,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType)>(myInfernalArrayV,4,amsoft::inutils::EShiftDirection::Left);
    amsoft::inutils::printArray(myInfernalArrayV,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType));
    amsoft::inutils::infernalShift<VeryElaboratedDataType,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType)>(myInfernalArrayV,10);
    amsoft::inutils::printArray(myInfernalArrayV,sizeof(myInfernalArrayV)/sizeof(VeryElaboratedDataType));

    return 0;
}

public static int[] solution1(int[] A, int K) {
    int temp[] = new int[A.length];

    int count = 0;

    int orignalItration = (K < A.length) ? K :(K%A.length); 


    for (int i = orignalItration; i < A.length; i++) {
        temp[i] = A[count++];
    }
    for (int i = 0; i < orignalItration; i++) {
        temp[i] = A[count++];
    }

    return temp;
}

#include <iostream>
#include <vector>

// same logic with STL implementation, but simpler, since no return value needed.
template <typename Iterator>
void rotate_by_gcd_like_swap(Iterator first, Iterator mid, Iterator last) {
    if (first == mid) return;
    Iterator old = mid;
    for (; mid != last;) {
        std::iter_swap(first, mid);
        ++first, ++mid;
        if (first == old) old = mid; // left half exhausted
        else if (mid == last) mid = old;
    }
}

int main() {
    using std::cout;
    std::vector<int> v {0,1,2,3,4,5,6,7,8,9};
    cout << "before rotate: ";
    for (auto x: v) cout << x << ' '; cout << '\n';
    int k = 7;
    rotate_by_gcd_like_swap(v.begin(), v.begin() + k, v.end());
    cout << " after rotate: ";
    for (auto x: v) cout << x << ' '; cout << '\n';
    cout << "sz = " << v.size() << ", k = " << k << '\n';
}

int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);

public int[] solution(int[] A, int K)
{
    for (var i = 0; i < gcd(A.Length, K); i++)
    {
        for (var j = i; j < A.Length - 1; j++)
        {
            var destIndex = ((j-i) * K + K + i) % A.Length;
            if (destIndex == i) break;
            var destValue = A[destIndex];
            A[destIndex] = A[i];
            A[i] = destValue;
        }
    }

    return A;
}

class Solution {
    public int[] solution(int[] A, int K) {
        // write your code in Java SE 8
        if (A.length > 0)
        {
            int[] arr = new int[A.length];
            if (K > A.length)
                K = K % A.length;

            for (int i=0; i<A.length-K; i++)
                arr[i+K] = A[i];

            for (int j=A.length-K; j<A.length; j++)
                arr[j-(A.length-K)] = A[j];

            return arr;
        }
        else
            return new int[0];
    }
}

int[] a  = {1,2,3,4,5,6};
    int k = 2;
    int[] queries = {2,3};

    int[] temp = new int[a.length];
    for (int i = 0; i<a.length; i++)
        temp[(i+k)%a.length] = a[i];

    a = temp;

# circle shift an array to the left by M 
def arrayCircleLeftShift(a, M):
    N = len(a)
    numAccessed = 0
    cycleIdx = 0
    while numAccessed != N:
        idx = cycleIdx
        swapIdx = (idx + M) % N
        tmp = a[idx]
        while swapIdx != cycleIdx:
            a[idx] = a[swapIdx]
            numAccessed += 1
            idx = swapIdx
            swapIdx = (idx + M) % N
        a[idx] = tmp
        numAccessed += 1
        cycleIdx += 1