移位数组元素

Question

我需要一些帮助，我知道之前有人问过这个问题，但我不明白也无法解决，所以我需要帮助。我需要将数组的元素移动到左侧的位置。因此，如果输入为 1、2、3、4、5，则输出将为 2、3、4、5、1。右边我也是这样，左边我搞不懂，还请解释一下逻辑，谢谢

    #include <iostream>
    using namespace std;
    int a[100],n,i,tempr,templ;
    int main()
    {
    cin>>n;
    for(i=1;i<=n;i++) cin >> a[i];
    for(i=1;i<=n;i++)
        {
            tempr = a[n];
            a[n] = a[i];
            a[i] = tempr;
            cout<<"Right: "<<a[i]<<endl;
        }
    for(i=1;i<=n;i++)
        {
            templ = a[2];
            a[2] = a[i];
            a[i] = templ;
            cout<<"Left: "<<a[i]<<endl;
        }
    return 0;
}

请帮忙！

Answer 1

在 C++ 中交换元素的最简单方法是使用 std::iter_swap()

所以对于一个包含 4 个元素的数组来交换元素 1 和 4，您可以执行以下操作

int a[4];
std::iter_swap(a, a+3);

请注意，您还需要 #include <algorithm> 才能正常工作

函数的基本逻辑是你给出2个元素在内存中的位置，所以数组的第一个元素也是它在内存中的位置，你可以传a + n，当n等于到要交换的元素的 n-1 索引号

Answer 2

Replace your code (to shift array left) with below code.

templ = a[0];
    for(i=0;i<n-1;i++)
    {
        a[i] = a[i+1];
        cout<<"Left: "<<a[i]<<endl;
    }
    a[n-1] = templ;
    cout<<"Left: "<<a[n-1]<<endl;

Answer 3

第一个问题是索引错误：

for(i=1;i<=n;i++) cin >> a[i]; //wrong logic, C++ indexing start from 0

正确做法：

for(i=0;i<n;i++) //all your loops

第二个问题是元素移位逻辑错误：更正版本：

//input example: 1 2 3 4 5
//to the left
int temp = a[0]; //remember first element
for(i=0;i<n-1;i++)
{
    a[i] = a[i+1]; //move all element to the left except first one
}
a[n-1] = temp; //assign remembered value to last element
//output: 2 3 4 5 1
cout << "To left: " << endl;
for(i=0;i<n;i++)
    cout << a[i] << endl;

//to the right
temp = a[n-1]; //remember last element
for(i=n-1;i>=0;i--)
{
    a[i+1] = a[i]; //move all element to the right except last one
}
a[0] = temp; //assign remembered value to first element
//output: 1 2 3 4 5 because elements are shifted back by right shift
cout << "To right: " << endl;
for(i=0;i<n;i++)
    cout << a[i] << endl;

编辑：

如何显示两个班次：

#include <iostream>
    using namespace std;
    int to_left[5], to_right[5],n,i,tempr,templ;
    int main()
    {

    cout << "Input array size: ";
    cin >> n;

    for(i=0;i<n;i++)
    {
        cin >> to_left[i]; //read values to first array
        to_right[i]=to_left[i]; //then copy values to second one
    }

    //shift first array to left
    int temp = to_left[0]; 
    for(i=0;i<n-1;i++)
    {
        to_left[i] = to_left[i+1]; //move all element to the left except first one
    }
    to_left[n-1] = temp; //assign remembered value to last element
    //output: 2 3 4 5 1
    cout << "To left: " << endl;
    for(i=0;i<n;i++)
        cout << to_left[i] << endl;

    //shift second array to right
    temp = to_right[n-1]; //remember last element
    for(i=n-1;i>=0;i--)
    {
        to_right[i+1] = to_right[i]; //move all element to the right except last one
    }
    to_right[0] = temp; //assign remembered value to first element
    //output: 1 2 3 4 5 because elements are shifted back by right shift
    cout << "To right: " << endl;
    for(i=0;i<n;i++)
        cout << to_right[i] << endl;

    return 0;
}

请注意，您的代码看起来非常像 C 代码。在 C++ 中，您可以在任何代码段中声明变量，而不仅仅是在开头。在 C++ 中，您可以像这样在 for 循环中声明变量： for(int i=0; i<...) - 不需要全局变量 i

作为参考，这将是一个很好的 C++ 代码示例，可以解决您面临的问题：

#include <iostream>
#include <vector>
int main()
{   
    std::size_t n; //size_t is unsiged type used for various sizes of containers or types
    std::cout << "Input array size: ";
    std::cin >> n;

    std::vector<int> to_left(n), to_right(n); //two dynamic arrays containing integers, takin n as their size

    for(std::size_t i=0;i<to_left.size();++i) //use vector size(), instead of n, also ++i in considered better for loops that i++ (may be faster)
    {
        std::cin >> to_left[i];
        to_right[i]=to_left[i];
    }

    int temp = to_left[0]; //declare temp here, not at the begining of code
    for(std::size_t i=0;i<n-1;++i)
        to_left[i] = to_left[i+1];
    to_left[n-1] = temp;

    std::cout << "To left: " << std::endl;
    for(std::size_t i=0;i<n;++i)
        std::cout << to_left[i] << std::endl;

    temp = to_right[n-1]; //reuse temp
    for(int i=to_right.size()-1;i>=0;--i) //note int, not std::size_t, because size_t is always >=0, loop would never end.
        to_right[i+1] = to_right[i];
    to_right[0] = temp;

    std::cout << "To right: " << std::endl;
    for(std::size_t i=0;i<n;i++)
        std::cout << to_right[i] << std::endl;

    return 0;
}

下面是理想的 C++ 代码：

#include <iostream>
#include <vector>
#include <algorithm>
int main()
{   
    std::size_t n;
    std::cout << "Input array size: ";
    std::cin >> n;

    std::vector<int> to_left(n), to_right(n);

    for(std::size_t i=0;i<to_left.size();++i)
    {
        std::cin >> to_left[i];
        to_right[i]=to_left[i];
    }

    // rotate first array to the left
    std::rotate(to_left.begin(), to_left.begin() + 1, to_left.end());

    // rotate second array to right
    std::rotate(to_right.rbegin(), to_right.rbegin() + 1, to_right.rend());

    std::cout << "To left:" << std::endl;
    for(auto x : to_left) //C++11 feature, x iterates through container
        std::cout << x << std::endl;

    std::cout << "To right:" << std::endl;
    for(auto x : to_right)
        std::cout << x << std::endl;

    return 0;
}

Answer 4

或者您可以使用 memmove(...) projected 正是为了这些目的，这里是您的示例：

#include <iostream> 
#include <cstring> 
using namespace std; 
//rotate Left
void r_left(int *a,int n) 
{ 
  int tmp=a[0];
  memmove(a,a+1,sizeof(int)*(n-1));
  a[n-1]=tmp;
} 
//rotate right
void r_right(int *a,int n) 
{ 
    int tmp=a[n-1];
    memmove(a+1,a,sizeof(int)*(n-1));
    a[0]=tmp;
 } 
void show(int *a,int n)
{
   while(n--)
   cout<<*a++<<' ';
   cout<<endl;
}

int main() 
{ 
   int ar[]={1,2,3,4,5};
   int n=sizeof(ar)/sizeof(ar[0]);
   r_left(ar,n);
   show(ar,n);
   r_right(ar,n);
   show(ar,n);
return 0;
}

Answer 5

    #include <iostream>
    using namespace std;
    int a[100], outR[100], outL[100], n, i;
    int main() {
      cin >> n;
      for (i = 0; i < n; i++) cin >> a[i];
      // Right
      for (i = 0; i < n; i++) {
        outR[i+1]= a[i];
      } 
      outR[0] = a[n-1]; // add first number
      // Left
      for (i = 1; i < n; i++) {
        outL[i-1]= a[i];
      }         
      outL[n-1] = a[0]; // add last number
      // Answer
      cout << "Right:\n";
      for(i=0; i<n; i++) {
        cout << outR[i] << endl;
        }
      cout << "Left:\n";
      for(i = 0; i < n; i++) {
        cout << outL[i] << endl;
      }
        return 0;
    }

简单的答案，您可以轻松看到所有内容，祝您好运。

您可能对 ,vector coding" 感兴趣，如果您花一些时间在这上面似乎会更容易：

    #include <iostream>
    #include <vector>
    using namespace std;
    vector <int> a, outR, outL;
    size_t i;
    int main () {
      int n, temp_int;
      cin >> n;
      while (n--) {
        cin >> temp_int;  // here you read number to your vector
        a.push_back(temp_int);  // here you add this to vector
        // remember that vector start from element 0 as like arrays
      }
      // Left
      // remember that last element will be first
      // you may have acces to size of your vector easily
      for (i = 0; i < (a.size()-1); i++) {
        outL.push_back(a.at(i+1));  // here you create new vector
      }
      outL.push_back(a.at(0)); // add last elemet which rotated

      // Right
      // to rotate left first you have push last element so
      outR.push_back(a.at(a.size()-1)); // add first elemet which rotated
      for (i = 1; i < a.size(); i++) {
        outR.push_back(a.at(i-1));  // here you push rest
        }

      cout << "Left"  << "\n";
      for (i = 0; i < a.size(); i++) {
        cout << outL.at(i) << endl;  // here you print value  
      }
      cout << "Right" << "\n";
      for (i = 0; i < a.size(); i++) {
        cout << outR.at(i) << endl;  // here you print value  
      }
      return 0;
    }

Answer 6

正如其他人已经指出的那样，一切都与指数有关。在 for 循环中，如果你的停止条件是 i <= size，你几乎总是有麻烦，因为 C++ 中的数组是零索引的。

Black Moses 算法是最容易理解的（可能也是最快的），我读你的代码就好像你试图通过数组交换数组的第一个值到最后一个位置。下面我试图指出这种方法。

#include <stdio.h>
#include <tchar.h>
#include <iostream>

void ShiftLeft(int* pArr, size_t length)
{
  for (size_t i = 1; i < length; i++)
  {
    int tmp = pArr[i - 1]; // Preserves the previous value
    pArr[i - 1] = pArr[i]; // Overwrites the previous position with the current value
    pArr[i] = tmp;         // Stores the previous value in the current position
                           // All in all the first value is swapped down the array until it is at the length - 1 position
                           // and all the other values are swapped to the left.

    /* For an array with 4 values the progression is as follows:
      i = 0: 1 2 3 4
      i = 1: 2 1 3 4
      i = 2: 2 3 1 4
      i = 3: 2 3 4 1    
    */

  }
}

void ShiftRight(int* pArr, size_t length)
{
  for (size_t i = length - 1; i > 0; i--)
  {
    // This code does exactly the same as for ShiftLeft but the loop is running backwards
    int tmp = pArr[i - 1];
    pArr[i - 1] = pArr[i];
    pArr[i] = tmp;
  }
}

void Print(int* pArr, size_t length)
{

  for (size_t i = 0; i < length; i++)
  {
    std::cout << pArr[i] << " ";
  }

  std::cout << std::endl;
}

int main()
{
  int arr[] = { 1, 2, 3, 4, 5, 6 };
  size_t length = sizeof(arr) / sizeof(arr[0]);

  Print(arr, length);
  ShiftLeft(arr, length);
  Print(arr, length);
  ShiftRight(arr, length);
  Print(arr, length);


  return 0;
}

Answer 7

int*  leftShiftOneByOneWIthoutTemp(int arr[], int sz)
{

    for (int i=0 ;i < sz-1; i++)
    {
        arr[i] = arr[sz-1] + arr[i];
        arr[sz-1] = arr[i] - arr[sz-1] ;
        arr[i] = arr[i] - arr[sz-1] ;
        std::cout << "iter  "<< i << std::endl;
        printArray(arr,5);

    }   
    std::cout << "final "<< std::endl;
    printArray(arr,5);

    return arr;
}

移位数组元素

Shift array elements

c++

arrays

algorithm

shift