C++ 反转数组中的内容

Reverse Contents in Array

提问by John

I have an array of numbers that I am trying to reverse. I believe the function in my code is correct, but I cannot get the proper output.

我有一组我想反转的数字。我相信我的代码中的函数是正确的，但我无法获得正确的输出。

The output reads: 10 9 8 7 6. Why can't I get the other half of the numbers? When I remove the "/2" from count, the output reads: 10 9 8 7 6 6 7 8 9 10

输出为：10 9 8 7 6. 为什么我不能得到另一半的数字？当我从计数中删除“/2”时，输出显示为：10 9 8 7 6 6 7 8 9 10

void reverse(int [], int);

int main ()
{
   const int SIZE = 10;
   int arr [SIZE] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

   reverse(arr, SIZE);
   return 0;
}
void reverse(int arr[], int count)
{
   int temp;
   for (int i = 0; i < count/2; ++i)
   {
      arr[i] = temp;
      temp = arr[count-i-1];
      arr[count-i-1] = arr[i];
      arr[i] = temp;

      cout << temp << " ";
   }
}

回答by juanchopanza

This would be my approach:

这将是我的方法：

#include <algorithm>
#include <iterator>

int main()
{
  const int SIZE = 10;
  int arr [SIZE] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
  std::reverse(std::begin(arr), std::end(arr));
  ...
}

回答by simonc

The line

线

arr[i] = temp;

is wrong. (On the first iteration of your loop it sets arr[i]to an undefined value; further iterations set it to an incorrect value.) If you remove this line, your array should be reversed correctly.

是错的。（在循环的第一次迭代中，它设置arr[i]为未定义的值；进一步的迭代将其设置为不正确的值。）如果删除此行，则数组应该正确反转。

After that, you should move the code which prints the reversed array into a new loop which iterates over the whole list. Your current code only prints the first count/2elements.

之后，您应该将打印反向数组的代码移动到一个新循环中，该循环遍历整个列表。您当前的代码仅打印第一个count/2元素。

int temp, i;
for (i = 0; i < count/2; ++i) {
    temp = arr[count-i-1];
    arr[count-i-1] = arr[i];
    arr[i] = temp;
}
for (i = 0; i < count; ++i) {
    cout << arr[i] << " ";
}

回答by Matt

Both answers look correct to me.

这两个答案在我看来都是正确的。

1. The first arr[i] = temp;should be removed

2. You should do a second loop to print allelements, not just half the array. The loop that does the reverse doesn't need to print it.

1. 第一个arr[i] = temp;应该删除

2. 您应该执行第二个循环来打印所有元素，而不仅仅是数组的一半。执行相反操作的循环不需要打印它。

回答by Floris

You are not printing the array, you are printing the value of temp- which is only half the array...

您不是在打印数组，而是在打印的值temp- 这只是数组的一半......

回答by user93353

void reverse(int [], int);
void printarray(int [], int );
int main ()
{
    const int SIZE = 10;
    int arr [SIZE] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

    cout<<"Before reverse\n";
    printarray(arr, SIZE);
    reverse(arr, SIZE);
    cout<<"After reverse\n";
    printarray(arr, SIZE);

    return 0;
}

void printarray(int arr[], int count)
{
    for(int i = 0; i < count; ++i)
        cout<<arr[i]<<' ';

    cout<<'\n';
}

void reverse(int arr[], int count)
{
   int temp;
   for (int i = 0; i < count/2; ++i)
   {
      temp = arr[i];
      arr[i] = arr[count-i-1];
      arr[count-i-1] = temp;
   }
}

回答by user93353

The solution to this question is very easy: Vectors

这个问题的答案很简单：向量

std::vector<int> vector;
for(int i = 0; i < 10;i++)
{
    vector.push_back(i);
}
std::reverse(vector.begin(), vector.end());

Voila! You are done! =)

瞧！你完成了！=)

Solution details:

解决方案详情：

This is the most efficent solution: Swap can't swap 3 values but reverse definitely can. Remember to include algorithm. This is so simple that the compiled code is definitely not needed.

这是最有效的解决方案：Swap 不能交换 3 个值，但 reverse 绝对可以。请记住包含算法。这太简单了，编译后的代码肯定是不需要的。

I think this solves the OP's problem

我认为这解决了 OP 的问题

If you think there are any errors and problems with this solution please comment below

如果您认为此解决方案有任何错误和问题，请在下方评论

回答by Argento

As a direct answer to your question: Your swapping is wrong

作为对您问题的直接回答：您的交换是错误的

void reverse(int arr[], int count){
   int temp;
   for(int i = 0; i < count/2; ++i){
      arr[i] = temp; // <== Wrong, Should be deleted
      temp = arr[count-i-1];
      arr[count-i-1] = arr[i];
      arr[i] = temp;
    }
}

assigning arr[i] = tempcauses error when it first enters the loop as temp initially contains garbage data and will ruin your array, remove it and the code should work well.

arr[i] = temp当它第一次进入循环时，分配会导致错误，因为 temp 最初包含垃圾数据并且会破坏你的数组，删除它并且代码应该可以正常工作。

As an advice, use built-in functions whenever possible:

作为建议，请尽可能使用内置函数：

• In the swapping you could just use swaplike std::swap(arr[i], arr[count-i-1])
• For the reverse as a whole just use reverselike std::reverse(arr, arr+count)

• 在交换中，您可以像使用交换一样 std::swap(arr[i], arr[count-i-1])
• 对于整个反向，只需使用reverselikestd::reverse(arr, arr+count)

I am using C++14 and reverse works with arrays without any problems.

我正在使用 C++14 并且反向使用数组没有任何问题。

回答by Abraham Hernandez

I would use the reverse()function from the <algorithm>library.

我会使用库中的reverse()函数<algorithm>。

Run it online: repl.it/@abranhe/Reverse-Array

在线运行：repl.it/@abranhe/Reverse-Array

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
  int arr [10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

  reverse(begin(arr), end(arr));

  for(auto item:arr)
  {
    cout << item << " ";
  }
}

Output:

输出：

10 9 8 7 6 5 4 3 2 1

Hope you like this approach.

希望你喜欢这种方法。

回答by thebarylowi

First of all what value do you have in this pice of code? int temp;? You can't tell because in every single compilation it will have different value - you should initialize your value to not have trash value from memory. Next question is: why you assign this temp value to your array? If you want to stick with your solution I would change reverse function like this:

首先，你在这段代码中有什么价值？int temp;? 您无法判断，因为在每次编译中它都会具有不同的值 - 您应该初始化您的值，以免内存中出现垃圾值。下一个问题是：为什么要将此临时值分配给数组？如果你想坚持你的解决方案，我会像这样改变反向功能：

void reverse(int arr[], int count)
{
    int temp = 0;
    for (int i = 0; i < count/2; ++i)
    {
        temp = arr[count - i - 1];
        arr[count - i - 1] = arr[i];
        arr[i] = temp;
    }

    for (int i = 0; i < count; ++i)
    {
        std::cout << arr[i] << " ";
    }
}

Now it will works but you have other options to handle this problem.

现在它可以工作了，但您还有其他选择来处理这个问题。

Solution using pointers:

使用指针的解决方案：

void reverse(int arr[], int count)
{
    int* head = arr;
    int* tail = arr + count - 1;
    for (int i = 0; i < count/2; ++i)
    {
        if (head < tail)
        {
            int tmp = *tail;
            *tail = *head;
            *head = tmp;

            head++; tail--;
        }
    }

    for (int i = 0; i < count; ++i)
    {
        std::cout << arr[i] << " ";
    }
}

And ofc like Carlos Abraham says use build in function in algorithmlibrary

像卡洛斯亚伯拉罕这样的ofc说在algorithm库中使用内置函数

回答by rashedcs

Procedure :

 1.Take an array.

 2.Then by default function reverse(array_name, array_name + size) .
  reverse(array_name, array_name + size) function exits in algorithm.h header file.

 3.Now print the array. 

 N.B  Here we use new and delete for dynamic memory allocation.

---

---

C++ implementation :

C++ 实现：

---

---

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


int main()
{
   int n;
   cin>>n;

   int *arr = new int[n];


   for(int i=0; i<n; i++)  cin>>arr[i];

   reverse(arr, arr+n);

   for(int i=0; i<n; i++)    cout<<arr[i]<<" ";

   delete[] arr;

   return 0;
}