阵列旋转 - 编程之家

如何解决阵列旋转

所以我想编写一个 C++ 程序来执行 r 次数组旋转。现在一次数组旋转意味着数组的第一个元素移动到最后一个位置，第 (i) 个位置的所有其他元素移动到第 (i-1) 个位置。

我面临的问题是，即使我已经为旋转应该发生的次数应用了 forloop，但数组也只旋转了一次。请帮助我解决什么问题。

#include<iostream>
using namespace std;
int main()
{
int i,n,r,arr1[100],arr2[100];
cout<<"Please enter the number of elements in your array: "<<endl;
cin>>n;
cout<<"Please enter the number of rotations to be performed: "<<endl;
cin>>r;
for(i=0;i<n;i++)
{
    cin>>arr1[i];
}    
for(i=0;i<r;i++)
{   
    arr2[n-1]=arr1[0];
    for(i=1;i<n;i++)
    {
       arr2[i-1]=arr1[i];
    }
}
cout<<"Printing Now:"<<endl;
for(i=0;i<n;i++)
{
    cout<<arr2[i]<<"\t";
}
}

解决方法

void Rotate(int arr[],int n,int r){
        /* To get the starting point of rotated array */
        int mod = r % n;
     
        // Prints the rotated array from start position
        for (int i = 0; i < n; i++)
            cout << (arr[(mod + i) % n]) << " ";
     
        cout << "\n";
    }

试试这个代码。您可以阅读有关此here

的更多信息 ,

对于想要完整工作代码的人：

#include <bits/stdc++.h>
using namespace std;


void leftRotate(int arr[],int k)
{
/* To get the starting point of rotated array */
int mod = k % n;

// Prints the rotated array from start position
for (int i = 0; i < n; i++)
    cout << (arr[(mod + i) % n]) << " ";

cout << "\n";
}

int main()
{
int n,k,arr[100];
cout<<"Please enter the number of elements: "<<endl;
cin>>n;
cout<<"Now enter the elements:"<<endl;
for(int i=0;i<n;i++)
{
    cin>>arr[i];
}
cout<<"Enter the number of rotations to be performed: "<<endl;
cin>>k;
  

leftRotate(arr,n,k);

return 0;
}

#include <bits/stdc++.h>
#define ll long long int

using namespace std;

int main()
{
    ll n,r;
    cin>>n>>r;
    ll a[n];

    /* pre-compute the index where the element would be present after rotation 
       by 'r' times and then insert it in the array */
    for(ll i=0; i<n; i++)
    {
        ll index = ((n+i)-r)%n;
        cin>>a[index];
    }

    // print the rotated array
    for(ll i=0; i<n; i++)
    {
        cout<<a[i]<<" ";
    }
}